course phy 232 |oԡpȒassignment #009
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18:18:34 Query introductory problem set 54 #'s 14-18. Explain whether, and if so how, the force on a charged particle due to the field between two capacitor plates is affected by its velocity.
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RESPONSE --> The field beteen the two plate will exert the same force on the charged particle no matter the velocity becaus ethe force between the two plates is constant.
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18:19:07 ** There is a force due to the electric field between the plates, but the effect of an electric field does not depend on velocity. The plates of a capacitor do not create a magnetic field. **
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18:21:00 Explain whether, and if so how, the force on a charged particle due to the magnetic field created by a wire coil is affected by its velocity.
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RESPONSE --> The force of the magnetic field on the wire is not due to the velocity but the distance between the charge and the field and the coil current and length.
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18:23:44 ** A wire coil does create a magnetic field perpendicular to the plane of the coil. If the charged particle moves in a direction perpendicular to the coil then a force F = q v B is exerted by the field perpendicular to both the motion of the particle and the direction of the field. The precise direction is determined by the right-hand rule. **
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RESPONSE --> The velocity does determine the force if the particle is traveling perpendicular to the coil then F= q v B.
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18:32:00 Explain how the net force changes with velocity as a charged particle passes through the field between two capacitor plates, moving perpendicular to the constant electric field, in the presence of a constant magnetic field oriented perpendicular to both the velocity of the particle and the field of the capacitor.
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RESPONSE --> The net force will be constant and equal to the sum of the force due to the magnetic field and due to the electric field. Since the velocity is normal to both, it is independent.
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18:58:29 ** At low enough velocities the magnetic force F = q v B is smaller in magnitude than the electrostatic force F = q E. At high enough velocities the magnetic force is greater in magnitude than the electrostatic force. At a certain specific velocity, which turns out to be v = E / B, the magnitudes of the two forces are equal. If the perpendicular magnetic and electric fields exert forces in opposite directions on the charged particle then when the magnitudes of the forces are equal the net force on the particle is zero and it passes through the region undeflected. **
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RESPONSE --> The net force will vary depending on the velocity because of the magnetic field but not the electric field.For the electric field, F=q E where the magnetic force is F= q v B. The forces are equal when v = E/B. The net force is 0 is the electric field and magnetic field have equal and opposite magnitude forces acting on the velocity which means no deflection.
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20:01:01 Query Principles and General Physics 20.2: Force on wire of length 160 meters carrying 150 amps at 65 degrees to Earth's magnetic field of 5.5 * 10^-5 T.
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RESPONSE --> The force is equal to F=I L B sin(theta). F= 160*150*5.5*10^5*sin65 F= 1.19 N
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20:02:08 The force on a current is I * L * B sin(theta) = 150 amps * 160 meters * 5.5 * 10^-5 T * sin(65 deg) = 1.20 amp * m * (N / (amp m) ) = 1.20 Newtons. Note that a Tesla, the unit of magnetic field, has units of Newtons / (amp meter), meaning that a 1 Tesla field acting perpendicular to a 1 amp current in a carrier of length 1 meters produces a force of 1 Newton. The question didn't ask, but be sure you know that the direction of the force is perpendicular to the directions of the current and of the field, as determined by the right-hand rule (fingers in direction of current, hand oriented to 'turn' fingers toward field, thumb in direction of force).
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20:04:41 Query Principles and General Physics 20.10. Force on electron at 8.75 * 10^5 m/s east in vertical upward magnetic field of .75 T.
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RESPONSE --> F= q v b F= 1.6 * 10^-19 C *8.75 * 10^5 m/s * .75T= 1.05E-13 N
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20:06:35 The magnitude of the force on a moving charge, exerted by a magnetic field perpendicular to the direction of motion, is q v B, where q is the charge, v the velocity and B the field. The force in this case is therefore F = q v B = 1.6 * 10^-19 C * 8.75 * 10^5 m/s * .75 T = 1.05 * 10^-13 C m/s * T = 1.05 * 10^-13 N. (units analysis: C m/s * T = C m/s * (N / (amp m) ) = C m/s * (kg m/s^2) / ((C/s) * m), with all units expressed as fundamental units. The C m/s in the numerator 'cancels' with the C m/s in the denominator, leaving kg m/s^2, or Newtons). The direction of the force is determined by the right-hand rule (q v X B) with the fingers in the direction of the vector q v, with the hand oriented to turn the fingers toward the direction of B. The charge q of the electron is negative, so q v will be in the direction opposite v, to the west. In order for the fingers to 'turn' qv toward B, the palm will therefore be facing upward, the fingers toward the west, so that the thumb will be pointing to the north. The force is therefore directed to the north.
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20:22:02 Query General Physics Problem (formerly 20.32, but omitted from new version). This problem is not assigned but you should solve it now: If an electron is considered to orbit a proton in a circular orbit of radius .529 * 10^-10 meters (the electron doesn't really move around the proton in a circle; the behavior of this system at the quantum level does not actually involve a circular orbit, but the result obtained from this assumption agrees with the results of quantum mechanics), the electron's motion constitutes a current along its path. What is the field produced at the location of the proton by the current that results from this 'orbit'? To obtain an answer you might want to first answer the two questions: 1. What is the velocity of the electron? 2. What therefore is the current produced by the electron? How did you calculate the magnetic field produced by this current?
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RESPONSE --> The fieldacting on the electron will be equal to the field produced by the proton at the radial distance. E= k* q / r. E= 27.22 N/C The force acting on the electron which is equal to the centripetal force is F= q E =4.56*10^18N The centipetal acc is F/m =a = 4.84*10^12 m/s^2. a= v^2 /r a r = v^2 4.84*10^12 * .529*10^-10 = v^2 v= 256 m/s F= q v B B= 6.54*10^-22 Teslas
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20:46:29 **If you know the orbital velocity of the electron and orbital radius then you can determine how long it takes to return to a given point in its orbit. So the charge of 1 electron 'circulates' around the orbit in that time interval. Current is charge flowing past a point / time interval. Setting centripetal force = Coulomb attraction for the orbital radius, which is .529 Angstroms = .529 * 10^-10 meters, we have m v^2 / r = k q1 q2 / r^2 so that v = sqrt(k q1 q2 / (m r) ). Evaluating for k, with q1 = q2 = fundamental charge and m = mass of the electron we obtain v = 2.19 * 10^6 m/s. The circumference of the orbit is `dt = 2 pi r so the time required to complete an orbit is `dt = 2 pi r / v, which we evaluate for the v obtained above. We find that `dt = 1.52 * 10^-16 second. Thus the current is I = `dq / `dt = q / `dt, where q is the charge of the electron. Simplifying we get I = .00105 amp, approx.. The magnetic field due to a .00105 amp current in a loop of radius .529 Angstroms is B = k ' * 2 pi r I / r^2 = 2 pi k ' I / r = 12.5 Tesla. **
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RESPONSE --> The equation k q1 q2 / r^2 equals E not force so how can it be used to equal the centripetal force.
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22:41:23 query univ 27.60 (28.46 10th edition). cyclotron 3.5 T field.
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RESPONSE --> The force of the field acting on the proton is F= q v B which is equal to F=m v^2 /r. When ke = .5 m v^2 v=3.25*10^33 m/s when m=1.66*10^-27 kg and ke =2.7MeV. m v^2 /r = q v B r =2.96*10^-9 m v= r w angular velocity = v/r = 1.09*10^42 rad/s
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22:52:24 What is the radius of orbit for a proton with kinetic energy 2.7 MeV?
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RESPONSE --> I just skipped the crtique of the previous problem by accident. KE= .5 m v^2 ke/ .5 m = v^2 2.7*10^6 / .5* 1.66*10^-27 = v^2 v=1.8^10^17 m/s m v^2 /r = q v b 1/r = q v b / m v^2 r= 1.87b Gm b cannot be found From here the problem cannot be solve because there is no b
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22:57:12 ** We know that the centripetal force for an object moving in a circle is F = m v^2 / r. In a magnetic field perpendicular to the velocity this force is equal to the magnetic force F = q v B. So we have m v^2 / r = q v B so that r = m v / (q B). A proton with ke 2.7 MeV = 2.7 * 10^6 * (1.6 * 10^-19 J) = 3.2 * 10^-13 J has velocity such that v = sqrt(2 KE / m) = sqrt(2 * 3.2 * 10^-13 J / (1.67 * 10^-27 kg) ) = 2.3 * 10^7 m/s approx.. So we have r = m v / (q B) = 1.67 * 10^-27 kg * 2.3 * 10^7 m/s / (1.6 * 10^-19 C * 3.5 T) = .067 m approx. **
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RESPONSE --> I didn't realize this was a continued problem. B is given as 3.5 T. The answer I recieved was right I believe but because I kept the units in eV the results were wrong and units did not match up. Other than that the work is correct in procedure.
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23:12:21 What is the radius of orbit for a proton with kinetic energy 5.4 MeV?
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RESPONSE --> The radius of orbit should be proportional to the other radius but but larger because the velocity must get larger due to the KE= .5 m v^2 We know that the centripetal force for an object moving in a circle is F = m v^2 / r. In a magnetic field perpendicular to the velocity this force is equal to the magnetic force F = q v B. So we have m v^2 / r = q v B so that r = m v / (q B). A proton with ke 5.4 MeV = 5.4 * 10^6 * (1.6 * 10^-19 J) = 8.64 * 10^-13 J has velocity such that v = sqrt(2 KE / m) = sqrt(2 * 3.2 * 10^-13 J / (1.67 * 10^-27 kg) ) = 1.03 * 10^15 m/s approx..
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23:13:48 ** Doubling the KE of the proton increases velocity by factor sqrt(2) and therefore increases the radius of the orbit by the same factor. We end up with a radius of about .096 m. **
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RESPONSE --> I understand the concept that doubling the KE of the proton increases velocity by factor sqrt(2) and therefore increases the radius of the orbit by the same factor but I followed the same formula but ended up with a much greater result and even tested many times.
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23:17:10 query univ 28.73 (was 28.52) rail gun bar mass m with current I across rails, magnetic field B perpendicular to loop formed by bars and rails What is the expression for the magnitude of the force on the bar, and what is the direction of the force?
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RESPONSE --> The expression used here is F= I L B where rather then refering to a charge the force the magnetic field applies is due to the current and length of the bar.
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23:25:34 ** The length of the bar is given as L. So the force is I L B, since the current and field are perpendicular. The acceleration of the bar is therefore a = I L B / m. If the distance required to achieve a given velocity is `ds and initial velocity is 0 then vf^2 = v0^2 + 2 a `ds gives us ds = (vf^2 - v0^2) / (2 a) = vf^2 / (2 a). If v stands for the desired final velocity this is written `ds = v^2 / (2 a). In terms of I, L, B and m we have `ds = v^2 / (2 I L B / m) = m v^2 / (2 I L B). Note that we would get the same expression using KE: since (neglecting dissipative losses) we have `dKE = `dW = F `ds we have `ds = `dKE / F = 1/2 m v^2 / (I L B). For the given quantities we get `ds = 25 kg * (1.12 * 10^4 m/s)^2 / (2 * 2000 amps * .50 Tesla * .5 meters) = 3.2 * 10^6 meters, or about 3200 km. **
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RESPONSE --> I didn't understand what was being asked for. The initial exquation is correct. In order to create an expression that related the mass projectable to the magnetic force KE, the acceleraction and velocity must be found.By using kinematics, the distance can be determined by varible representation of the possible final velocity which is dependent on the KE and the magnetic force. The acceleration of the bar is therefore a = I L B / m. If the distance required to achieve a given velocity is `ds and initial velocity is 0 then vf^2 = v0^2 + 2 a `ds gives us ds = (vf^2 - v0^2) / (2 a) = vf^2 / (2 a). If v stands for the desired final velocity this is written `ds = v^2 / (2 a). In terms of I, L, B and m we have `ds = v^2 / (2 I L B / m) = m v^2 / (2 I L B).
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23:36:40 query 28.66 u quark + 2/3 e and d quark -1/3 e counterclockwise, clockwise in neutron (r = 1.20 * 10^-15 m) What are the current and the magnetic moment produced by the u quark?
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RESPONSE --> Assuming the quark has 1/3 the mass of the neutron. F= m v^2 /r = q v b The u quark has half the charge of the neutron and will therefore contribute to have of the magnetic moment when it is 0. From this point I'm not sure of the way to answer the problem. It doesn't seem like there is enough information.
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23:45:00 ** If r is the radius of the orbit and v the velocity then the frequency of an orbit is f = v / (2 pi r). The frequency tells you how many times the charge passes a given point per unit of time. If the charge is q then the current must therefore be }I = q f = q v / (2 pi r). Half the magnetic moment is due to the u quark, which carries charge equal and opposite to the combined charge of both d quarks, the other half to the d quarks (which circulate, according to this model, in the opposite direction with the same radius so that the two d quarks contribute current equal to, and of the same sign, as the u quark). The area enclosed by the path is pi r^2, so that the magnetic moment of a quark is I A = q v / (2 pi r) * pi r^2 = q v r / 2. The total magnetic moment is therefore 2/3 e * v r / 2 + 2 ( 1/3 e * v r / 2) = 4/3 e v r / 2 = 2/3 e v r.. Setting this equal to the observed magnetic moment mu we have 2/3 e v r = mu so that v = 3/2 mu / (e r) = 3/2 * 9.66 * 10^-27 A m^2 / (1.6 * 10^-19 C * 1.20 * 10^-15 m) = 7.5 * 10^7 m/s, approx.. Note that units are A m^2 / (C m) = C / s * m^2 / (C m) = m / s. **
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RESPONSE --> The current is the amount of charge past a point at a a point so by knowing that the particle travels in a circle it can be found that the frequency is v/ 2 pi r. The current is then the f*q. The moment due to u is half because it has half the charge, 9.66*10^-27 a m^2 divided by 2. The moment is then I * A. This can then be used to find the velocity.
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23:45:38 query univ 28.68 (29.56 10th edition) infinite L-shaped conductor toward left and downward. Point a units to right of L along line of current from left. Current I. What is the magnetic field at the specified point?
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RESPONSE --> It is hard to tell what the problem is saying because the numbers do not match up. The only part of the conductor that will affect point a is the part perpendicular to it. But I'm not sure how to calculate the distance between the point and the conductor.
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23:46:14 STUDENT RESPONSE FOLLOWED BY SOLUTION: I could not figure out the magnetic field affecting point P. the current is cursing ** I assume you mean 'coursing', though the slip is understandable ** toward P then suddnely turns down at a right angle. If I assume that the magnetic field of a thin wire is radial in all directions perpendicular to the wire, then it is possible that at least one field line would be a straight line from the wire to point P. It seems to me that from that field line,down the to the lower length of the wire, would affect at P. SOLUTION: The r vector from any segment along the horizontal section of the wire would be parallel to the current segment, so sin (theta) would be 0 and the contribution `dB = k ' I `dL / r^2 sin(theta) would be zero. So the horizontal section contributes no current at the point. Let the y axis be directed upward with its origin at the 'bend'. Then a segment of length `dy at position y will lie at distance r = sqrt(y^2 + a^2) from the point and the sine of the angle from the r vector to the point is a / sqrt(y^2 + a^2). The field resulting from this segment is therefore `dB = k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2). Crossing the I `dy vector with the r vector tells us the `dB is coming at us out of the paper (fingers extended along neg y axis, ready to 'turn' toward r results in thumb pointing up toward us away from the paper). This is the direction for all `dB contributions so B will have the same direction. Summing all contributions we have sum(k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2), y from 0 to -infinity). Taking the limit as `dy -> 0 we get the integral of k ' I a / (a^2 + y^2)^(3/2) with respect to y, with y from 0 to - infinity. This integral is -k ' I / a. So the field is B = - k ' I / a, directed upward out of the page. **
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RESPONSE --> The equation to be used here is dB = k ' I `dL / r^2 sin(theta) where theta is the angle of the of the field to the perpendicular of the wire. To find the distance, dB = k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2). Integrate from 0 to -infinity to get the field,B = - k ' I / a.
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Q]ƜŦC assignment #010 010. `Query 31 Physics II 06-29-2008
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12:41:02 Query Principles and General Physics 21.04. A circular loop of diameter 9.6 cm in a 1.10 T field perpendicular to the plane of the loop; loop is removed in .15 s. What is the induced EMF?
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RESPONSE --> the emf is equal to the area of the loop times and field acting through the loop times the angle times the time. EMF= .15*2*pi*4.8*1.1*cos(0)
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12:44:09 The average induced emf is the average rate of change of the magnetic flux with respect to clock time. The initial magnetic flux through this loop is flux = magnetic field * area = 1.10 T * (pi * .048 m)^2 = .00796 T m^2. The flux is reduced to 0 when the loop is removed from the field, so the change in flux has magnitude .0080 T m^2. The rate of change of
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RESPONSE --> This means that the induced emf is the change in flux over the change in time after calculating flux by flux = magnetic field * area = 1.10 T * (pi * .048 m)^2 = .00796 T m^2. The equal I used was for circumference and should have been for area. The flux goes form the above 0.08 to 0 divided by the tim .15s so .008/.15 .0533 = induced emf.
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12:44:17 flux is therefore .0080 T m^2 / (.15 sec) = .053 T m^2 / sec = .053 volts.
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12:48:11 query gen problem 21.23 320-loop square coil 21 cm on a side, .65 T mag field. How fast to produce peak 120-v output? How many cycles per second are required to produce a 120-volt output, and how did you get your result?
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RESPONSE --> I do not understand how to go form amount of flux and the magnetic field to calculate the voltage output. The flux = .65 * .21^2 * 320 = 9.1728 W.
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Ʃϼ 浃Kةk{ assignment #011 011. `query 1 Physics II 06-29-2008
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13:01:32 query introset 5 # 12: Finding the conductivity given rate of energy flow, area, temperatures, thickness of wall. Describe how we find the conductivigy given the rate of energy flow, area, temperatures, and thickness of the wall?
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RESPONSE --> Conductivity is equal to the rate of change in energy flow divided by the area times the temperature gradient. To find the temperature gradient, find the temperature change final minus initial and divide by the thickness of the wall. Then plug the values in to get conductivity.
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13:03:13 ** The rate at which thermal energy is conducted across for a object made of a given substance is proportional to the temperature gradient (the rate at which temperature changes across the object), and to the cross-sectional area of the object. The conductivity is the constant of proportionality for the given substance. So we have Rate of thermal energy condction = conductivity * temperature gradient * area, or R = k * `dT/`dx * A. For an object of uniform cross-section `dT is the temperature difference across the object and `dx is the distance L between the faces of the object. In this case the equation is R = k * `dT / L * A and we can solve to get k = R * L / (`dT * A). **
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RESPONSE --> The equation meantion is the same as the sybolic one here R = k * `dT/`dx * A.
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13:04:43 Explain how energy flow, for a given material, is affected by area (e.g., is it proportional to area, inversely proportional, etc.), thickness and temperature gradient.
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RESPONSE --> Energy flow is proportional to area and thickness and the temperature gradient. As each one of these values increase the energy flow also increases.
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13:09:00 ** CORRECT STUDENT ANSWER WITHOUT EXPLANATION: Energy flow is directly proportional to area inversely propportional to thickness and directly proportional to temperature gradient GOOD STUDENT ANSWER WITH EXPLANATIONS, PLUS INSTRUCTOR COMMENTARY: The energy flow for a given material increases if the area increases. This is because the more area you have the wider a path something has to go through so more of it can move through it. Just like a 4 lane highway will carry more cars in a given time interval than a two lane highway will. So the relationship of energy flow to area is proportional. Energy flow, however is inversely proportional to thickness. This is because although the thermal energy flows through the material, the material impedes it. So if the thickness increases the thermal energy will have to travel farther through the resistance and be impeded more. ** Also for given temperature difference, greater thickness `dx implies smaller temperature gradient `dT / `dx. Small temperature gradient doesn't 'drive' the energy flow as much. Energy flow is also proportional to the temperature gradient. Meaning if the difference in the two temperatures is greater then the energy will move faster from one side to the other. Temperature gradient is not difference in temperatures, it's difference in temperature per unit of distance across the material. Temperature gradient is `dT / `dx, not just `dT. Greater temperature gradient means greater difference in temperature over any given small distance increment. The greater the temperature difference across this increment the more energy will flow. **
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RESPONSE --> Energy flow is not proportional to thickness because greater thickness means that the energy must flow through the material that resists the energy movement so they are inversely proportional.
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13:10:32 principles of physics and general college physics 13.8: coeff of expansion .2 * 10^-6 C^-1, length 2.0 m. What is expansion along length if temp increases by 5.0 C?
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RESPONSE --> dL = coef * length * dT dL= 2*10^-6 m
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13:11:28 The amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference: `dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. This is 2 microns, two-onethousandths of a millimeter. By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1);using this for the coefficient of expansion yields a change in length of 24 * 10^-6 m, or 24 microns, which is 240 times as much as for the given alloy.
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13:12:09 ** The coefficient of volume expansion for quartz is 1 x 10^(-6) C^(-1). The sphere has diameter 8.75 cm, so its volume is 4/3 pi r^3 = 4/3 pi ( 4.38 cm)^3 = 352 cm^3, approx.. We therefore have dV = beta* V0*dT = 3 x 10^(-6) C^ (-1) * 352 cm^3 * (200C - 30 C) = 0.06 cm^3 **
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. k
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13:16:57 query univ 17.103 (15.93 10th edition) (Note that Principles of Physics and General College Physics students don't do University Physics problems).
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RESPONSE --> I have the 12th edition book and do not have a picture of the problem
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13:18:49 ** The ice doesn't change temperature until it's melted, at which time it is in the form of water with the specific heat of water. Also the steam will come to temperature Tf so its change in thermal energy after being condensted will be 4186 J / (kg K) * .035 kg * (Tf - 100 C). I prefer to say that the sum of all the thermal energy changes is zero, so that we don't have to worry about taking a negative of a negative (which you should have done on your right-hand side, and which would have avoided the negative result). I would write the equation as follows: [.446 kg * 390 J/kg*K * (Tf - 0 C)] + [0.0950 kg * 4186 J/kg*K *(Tf - 0 C)] - .0350 kg * 2.256 x 10^6 J/kg + 4186 J / (kg K) * .035 kg * (Tf - 100 C) = 0. Noting that change in temperature of a Kelvin degree is identical to a change of a Celsius degree this gives you 170 J/C * Tf + 390 J/C * Tf - 79000 J - 14000 J + 140 J / C * Tf = 0 or 700 J / C * Tf = 93000 J, approx. or Tf = 130 C. This isn't possible--we can't end up warmer than the original temperature of the steam. We conclude that not all the steam condenses and that the system therefore reaches equilibrium at 100 C, with a mixture of water and steam. Our energy conservation equation will therefore be [.446 kg * 390 J/kg*C * (100 C - 0 C)] + [0.0950 kg * 4186 J/kg*C *(100 C - 0 C)] - mCondensed * 2.256 x 10^6 J/kg = 0 where mCondensed is the mass of the condensed steam. This gives us 17000 J + 39000 J - mCondensed * 2.3 * 10^6 J/kg = 0 or mCondensed = 56000 J / (2.3 * 10^6 J/kg) = .023 kg. We end up with .095 kg * .023 kg = .118 kg of water and .035 kg - .023 kg = .012 kg of steam. **
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13:30:53 query univ phy 17.100 (90 in 10th edition): C = 29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T . Give your solution to this problem.
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RESPONSE --> dQ= nC dt C = 29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T Q= 3(29.5 T + (4.1+ 10^-3 J/mol K^2) T^2 from 27 to 227 degrees. Q= 20723.3067 J -2398.4667 = 18324.84 J
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13:32:25 ** Specific heat is not constant but varies with temperature. The energy required to raise the temperature of 3 moles by dT degrees while at temperature T is 3 mol * C * dT = 3 mol * (29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T) * dT. You have to integrate this expression from T= 27 C to T = 227 C, which is from 300 K to 500 K. Antiderivative of (29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T) is F(T) = 29.5 J / (mol K) * T + (8.2 + 10^-3 J/mol K^2) * T^2 / 2. Simplify and apply Fundamental Theorem of Calculus (find F(500) - F(300) if you think the temperature T is in Kelvin or F(227) - F(27) if you think it's in Celsius; this isn't specified in the problem and while the units tend to imply Kelvin temperature the resulting specific heats would be unrealistic for most real substances), then multiply by the constant 3 moles. The result for Kelvin temperatures is about 20,000 Joules. **
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RESPONSE --> My solution is correct put the units weren't changed to kelvins.
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13:47:41 **The final mass of the system is .525 kg, meaning that .525 kg - (.340 kg + .150 kg) = .035 kg of steam condensed then cooled to 71 C. The thermal energy change of the calorimeter plus the water is .150 kg * 420 J/(kg C) * 56 C + .34 kg * 4187 J / (kg C) * 56 C = 83,250 J, approx. The thermal energy change of the condensed water is -Hf * .035 kg + .035 kg * 4187 J / (kg C) * (-29 C) = -Hf * .035 kg - 2930 J, approx. Net thermal energy change is zero, so we have 83,250 J - Hf * .035 kg - 4930 J = 0 so that Hf = 79,000 J / (.035 kg) = 2,257,000 J / kg. **
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. Noooo. I skipped it. Well the cal. mass is .150 kg and water and cal. is .525kg at end. .375 lg is water that is vaporized. That means that an additional .035 kg of water in the air as condensed. The energy of the water is m C dT =79720.48 J + the energy of the cal. 3520 = 83248J Im not sure I understand how the condensed water calculation is made and how it gets to the heat of vaporization.
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Wˆx assignment #012 012. `query 2 Physics II 06-29-2008
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15:31:37 query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht Explain how the final temperature of the system is combined with the known initial temperatures and masses of both substances to obtain the unknown specific heat
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RESPONSE --> The units of specific heat are J/ kg* K so the energy is needed in J thatn changed the temperature divided bay the mass timesn the temperature change in kelvin.
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15:35:19 ** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as `dQ = mass * specific heat * `dT. (General College and University Physics students note that specific heat is not in fact strictly constant and for most substances changes with temperature.) We know that what energy one substance loses, the other gains, and we know the specific heat of water , so we can set the two quantities equal and solve for specific heat of the unknown substance. For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently m1 c1 `dT1 = - m2 c2 `dT2. If we know all the quantities except, say, the specific heat c2 of the second substance we can easily solve for c2. **
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RESPONSE --> The equation used to calculate the specific heat is `dQ = mass * specific heat * `dT where Q is the change in the thermal energy of the system. The dQ of both subtances added together should equal 0 since no energy is lost in isloated systems. `dQ1 + `dQ2 = 0, which gives us the equation m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently m1 c1 `dT1 = - m2 c2 `dT2. If we know all the quantities except, say, the specific heat c2 of the second substance we can easily solve for c2. **
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15:35:45 prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.
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RESPONSE --> q
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15:37:32 The Kelvin temperature is 273 above the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree). 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -(100 + 273) K = 373 K, (5500 + 273) K = 5773 K. 78 F is (78 F - 32 F) = 46 F above the freezing point of water. 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing, or since freezing is at 0 C, just 26 C. The Kelvin temperature is therefore (26 + 273) K = 299 K. -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing, which is -273 C or (-273 + 273) K = 0 K. This is absolute zero, to the nearest degree.
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RESPONSE -->
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15:42:48 prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm.
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RESPONSE --> PV=nRT n and R are constant and are not needed. P=(1/9 )(20) P= 2.22 40 = 1/9 T T= 360 C
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15:43:04 query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge
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RESPONSE -->
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15:43:11 ** T1 = 288 K, T2 = 311 K so T2 / T1 = 311 / 288 = 1.08, approx. This is approx. an 8% increase in temperature. The pressure will end up at 3ll / 288 * 321 kPa = 346 kPa, approx (note that we have to use actual rather than gauge pressure so init pressure is 220 kPa + 101 kPa = 321 kPa, approx. You then have to change the number n of moles of gas to get back to 331 kPa, so n3 / n2 = P3 / P2 = 321 kPa / 346 kPa or approximately .93, which would be about a 7% decrease. Note that the results here are mental estimates, which might not be particularly accurate. Work out the process to see how the accurate numbers work out. **
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16:04:25 query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area?
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RESPONSE --> H= A e d T^4 H= pi * 6.96 * 10^8 ^2 * 1 * 5.67*10^-8 * T^4 I'm not sure how to calculate the radiation if there is no temperature given.
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16:06:55 ** GOOD STUDENT SOLUTION The total radiation of the sun was the rate it reaches earth times the imaginary surface of the sphere from the sun center to earth atmosphere, or 1500 W/m^2 * (4`pir^2) = 1500W/m^2 * 2.8537 x10^23 m^2 = 4.28055 x 10 ^ 26 W. } Radiation per unit of area surface of the sun would be
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RESPONSE --> The area of the sphere used should be the surface area which is 4 pi r^2 and the rate is given times the area given the total radiation on the earth in W.
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16:12:28 4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2 If the sun is radiating as an ideal blackbody, e = 1, the T would be found: H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K) T^4 = 1.240 * 10 ^ 15 K ^4 T = 5934.10766 K on surface of sun. **
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RESPONSE --> The radiation divided by the surface area is the radiation per m^2 or w/m^2. A blackbody absorbs all radiation and emitts all so the emissivity is 1. Using the equation H = `dQ/`dt solving for the temperature is easy once the radiation total is found and solving for T gets 5900K
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16:15:30 univ phy 17.115 time to melt 1.2 cm ice by solar radiation 600 w/m^2, 70% absorption, environment at 0 C.
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RESPONSE --> cannot find problem
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16:25:01 ** Thermal energy is not radiating in significant quantities from the ice, so you use 70% of the incoming 600 watts/m^2, which gives you 420 watts / m^2, or 420 Joules/second for every square meter if ice. Melting takes place at 0 C so there is no thermal exchange with the environment. A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can correct this by using the correct density of ice). It takes about 330,000 Joules to melt a kg of ice, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec the time required will be about 10,000 seconds, or around 3 hours. All my calculations are approximate and done mentally so you should check them yourself, using more precise values of the constants, etc. **
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RESPONSE --> Only 70% of the enrgy is transferred so the 600watts becomes 420 watts/m^2. The ice only stops melting at below 0 C so when this is multplied by the area of the ice melting we get the amount of energy required over time.
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_˾ܾLғK| assignment #013 013. `query 3 Physics II 06-29-2008
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16:29:42 query problem 15 introductory problem sets temperature and volume information find final temperature. When temperature and volume remain constant what ratio remains constant?
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RESPONSE --> PV=nRT If volume and temperture are constant then the equation becomes P= nR for final and the initial values.
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16:31:05 ** PV = n R T so n R / P = T / V; since T and V remain constant T / V and therefore n R / P remain constant; since R is constant it follows that n / P remains constant. **
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RESPONSE --> Since R is always constant the n/P must also be constant or proportional in a way that they are always equal to R.
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16:32:51 why, in terms of the ideal gas law, is T / V constant when only temperature and volume change?
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RESPONSE --> I don't know how it is possible that temperature and volume can be constant and change at the same time. But when they are varing they vary dependent on each other so maybe taht is the reason.
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16:34:36 ** STUDENT ANSWER AND INSTRUCTOR RESPONSE: They are inverselt proportional. They must change together to maintain that proportion. INSTRUCTOR RESPONSE: You haven't justified your answer in terms of the ideal gas law: PV = n R T so V / T = n R / P. If only T and V change, n and P don't change so n R / P is constant. Therefore V / T is constant, and so therefore is T / V. You need to be able to justify any of the less general relationships (Boyle's Law, Charles' Law, etc.) in terms of the general gas law. **
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RESPONSE --> Since V/T is equal to the nR /P the later values never change so that means V/T is constant even though there individual values may not be.
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16:34:53 prin and gen problem 14.3: 2500 Cal per day is how many Joules? At a dime per kilowatt hour, how much would this cost?
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RESPONSE -->
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16:40:38 One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 13,000,000 Joules. A watt is a Joule per second, a kilowatt is 1000 Joules / second and a kiliowatt-hour is 1000 Joules / second * 3600 seconds = 3,600,000 Joules. 13,000,000 Joules / (3,600,000 Joules / kwh) = 4 kwh, rounded to the nearest whole kwh. This is about 40 cents worth of electricity. It's worth noting that you use 85% of this energy just keeping yourself warm, so the total amount of physical work you can produce in a day is worth less than a dime.
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RESPONSE --> 1 Cal is equal to 1 kilocalorie =4186J A watt is J/s a kilowatt-hour is 1000 Joules / second * 3600 seconds = 3,600,000 Joules/s 13MJ / 3.6MJ/s = 3.6Kwh
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17:24:36 prin phy and gen phy problem 14.07 how many Kcal of thermal energy would be generated in the process of stopping a 1200 kg car from a speec of 100 km/hr?
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RESPONSE --> i'M NOT SURE HOW TO GO ABOUT SOLVING FOR kCAL BUT i THINK THAT the amount of work required to stop the car should be found.
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17:30:31 **STUDENT SOLUTION for 1000 kg car at 100 km/hr WITH ERROR IN CONVERSION OF km/hr TO m/s: The book tells that according to energy conservation initial KE = final KE + heat or (Q) 100km/hr *3600*1/1000 = 360 m/s ** 100 km / hr is 100 * 1000 meters / (3600 sec) = 28 m/s or so. With units your conversion would be 100km/hr *3600 sec / hr *1/ (1000 m / km) = 360 km^2 sec / ( m hr^2). Correct conversion with conversion factors would be 100 km / hr * (1000 m / km) * (1 hr / (3600 sec) = 28 m/s, approx. Otherwise your solution is correct. STUDENT SOLUTION WITH DIFFERENT ERROR IN UNITS Ke=0.5(1000Kg)(100Km)^2 = 5MJ 1Kcal=4186J 5MJ/4186J==1194Kcal INSTRUCTOR COMMENT: Right idea but 100 km is not a velocity and kg * km^2 does not give you Joules. 100 km / hr = 100,000 m / (3600 sec) = 28 m/s, approx. so KE = .5(1000 kg)(28 m/s)^2 = 400,000 J (approx.). or 100 Kcal (approx). **
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RESPONSE --> I do not understand the explanation. Is the right answer even there?
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17:30:40 query gen phy problem 14.13 .4 kg horseshoe into 1.35 L of water in .3 kg iron pot at 20 C, final temp 25 C, spec ht of hossshoe
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RESPONSE -->
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17:39:26 ** STUDENT SOLUTION BY EQUATION (using previous version in which the amount of water was 1.6 liters; adjust for the present 1.35 liters): M1*C1*dT1 + M2*C2*dT2 = 0 is the equation that should be used. 0.4kg * 450J/Kg*Celcius *(25celcius -T1) + 0.3kg * 450J/kg*celcius * (25 celcius - 20 celcius) + 1.6kg *4186J/kg*celcius * (25 celcius - 20 celcius) = 0 } Solve for T1, T1 = 214.8 Celsius Solution below is 189.8 C. GOOD STUDENT SOLUTION: This problem is very similar to the first video experiment. You can determine in the same way the thermal energy gained by the water. The pot gains a little thermal energy too,and you have to know how many Joules are gained by a kg of the iron per degree per kg, which is the specific heat of iron (give in text). You can therefore find the energy lost by the horseshoe, from which you can find the energy lost per degree per kg. For this problem I think we are assuming that none of the water is boiled off ass the hot horse shoe is dropped into the water. First I will find the initial temperature of the shoe. Since water's density is 1g/ml, 1 milliliter of water will weigh 1 gram. So 1.35 Liters of water will have a mass of 1.35 kg. 1.35kg of water is heated by 5 degrees The specific heat of water is 4186 J/kg/degrees celsius so 4186 J / kg / C *1.35 kg * 5 C = 28255 J is necessary to heat the water but since it is in equilibrium with the bucket it must be heated too. mass of bucket = 0.30 kg specific heat of iron = 450 J/kg/degrees 450 J / kg / C * 0.30 kg * 5 C = 675 J to heat bucket So it takes 675 J to heat bucket to 25 degrees celsius 28255 J to heat water to 25 degrees celsius so the horse shoe transferred 675+28255 = 28930 J of energy. Mass of horse shoe = 0.40 kg horse shoe is also iron specific heat of iron = 450 J/kg/degree 28930 J / 0.40kg =72,326 J / kg 72,326 J/kg / (450 J / kg / C) = 160.7 C, the initial temperature of the horseshoe. STUDENT ERROR: MISSING COMMON KNOWLEDGE: Estimate 1.60L of water = 1KG. Could not find a conversion for this anywhere. 1 liter = 1000 mL or 1000 cm^3. Density of water is 1 gram/cm^3 so 1 liter contains 1000 g = 1 kg. Alternatively, density of water is 1000 kg / m^3 and 1 liter = .001 m^3, leading to same conclusion. **
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17:51:47 query univ problem 18.60 (16.48 10th edition) 1.5 L flask, stopcock, ethane C2H6 at 300 K, atm pressure. Warm to 380 K, open system, then close and cool. What is the final pressure of the ethane and how many grams remain? Explain the process you used to solve this problem.
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RESPONSE --> I assumed I did have to do the previous no univ. problems. P1 V1/nRT1 = P2 V2/nRT2 the volume, moles and constant R are all constant and can be ignored. P1 / T1 = P2 / T3 P2= P1 T3 / T1 = 128313 Pa or 1.28*10^5 Pa
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17:57:55 ** use pV = nRT and solve for n n = p V / (r T) = (1.283 *10^5 Pa )(1.50 * 10^-3 m^3 ) / [ (8.36 J / (mol K) )(380 K) ] = .062 mol, approx.. The atomic masses of 2 C and 6 H add up to 30.1, meaning 30.1 grams / mol. So total mass of the gas is initially m(tot) = (.062 mol)(30.1 g/mol) m(tot) = 1.86 g Now if the system is heated to 380 K while open to the atmosphere, pressure will remain constant so volume will be proportional to temperature. Therefore the volume of the gas will increase to V2 = 1.5 liters * 380 K / (300 K) = 1.9 liters. Only 1.5 liters, with mass 1.5 liters / (1.9 liters) * 1.86 grams = 1.47 grams, will stay in the flask. The pressure of the 1.47 grams of ethane is 1 atmosphere when the system is closed, and is at 380 K. As the temperature returns to 300 K volume and quantity of gas will remain constant so pressure will be proportional to temperature. Thus the pressure will drop to P3 = 1 atm * 300 K / (380 K) = .79 atm, approx.. **
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RESPONSE --> I'm confused Why that volume changes during the heating I gues it is because the container is open but I would have thought that the weight and volume would remain the same because extra would flow out of the system. Then higher temperature would mean higher pressure but I guess I am wrong. I do not understand whats going on.
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18:02:02 univ phy query problem 18.62 (16.48 10th edition) unif cylinder .9 m high with tight piston depressed by pouring Hg on it. How high when Hg spills over? How high is the piston when mercury spills over the edges?
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RESPONSE --> I do not see this problem in the book, 18.62 is similar but has not Hg on it so I'm confused because the information given is not enough to solve the problem. There should be a radius given i think.
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18:04:33 ** Let y be the height of the mercury column. Since T and n for the gas in the cylinder remain constant we have P V = constant, and since cross-sectional area remains constant V = A * h, where h is the height of the air column, we have P * h = constant. Thus P1 h1 = P2 h2, with P1 = atmospheric pressure = Patm and h1 = .9 m, P2 = Patm + rho g y. Mercury spills over when the depth of the mercury plus that of the air column is .9 m, at which point h2 = h1 - y. So the equation becomes Patm * h1 = (Patm + rho g y) * (h1 - y). We can solve this equation for y (the equation is quadratic). We obtain two solutions; one y = 0 tells us what when there is no mercury (y = 0) there is no deflection below the .9 m level. The other solution is y = (gh1rho - Pa)/(grho) = .140 m, which tells us that .140 m of mercury will again bring us to .9 m level. We might assume that this level must correspond to the level at which mercury begins spilling over. To completely validate this assumption we need to show that the level of the top of the column will be increasing at this point: The level of the top of the mercury column above the bottom of the cylinder can also be regarded as a function f (y) of the depth of the mercury. If mercury depth is y then the pressure in the cylinder is Patm + rho g y and the height of the gas in the cylinder is Patm / (Patm + rho g y ) * h1. The level of the mercury is therefore f(y) = Patm / (Patm + rho g y) * h1 + y The derivative of this function is f ' ( y ) = 1 - Patmgh1rho/(grhoy + Patm)^2, which is a quadratic function of y. Multiplying both sides by (rho g y + Patm)^2 we solve for y to find that y = sqrt(Patm)(sqrt(g* h1 * rho) - sqrt(Patm) )/ (grho) = .067 m approx., is a critical point of f(y). The second derivative f '' (y) is 2 Patmg^2h1rho^2/(grhoy + Patm)^3, which is positive for y > 0. This tells us that any critical point of f(y) for which y > 0 will be a relative minimum. So for y = .0635 m we have the minimum possible total altitude of the air and mercury columns, and for any y > .0635 m the total altitude is increasing with increasing y. This proves that at y = .140 m the total height of the column is increasing and additional mercury will spill over. To check that y = .140 m results in a total level of .9 m we note that the air column would then be .9 m - .140 m = .760 m, resulting in air pressure .9 / .760 * 101300 Pa = 120,000 Pa. The pressure due to the .140 m mercury column is 19,000 Pa, which when added to the 101,300 Pa of atmospheric pressure gives us 120,000 Pa, accurate to 3 significant figures. The gauge pressure will be 19,000 Pa. A more direct but less rigorous solution: The cylinder is originally at STP. The volume of the air in the tube is inversely proportional to the pressure and the altitude of the air column is proportional to the volume, so the altitude of the air column is inversely proportional to the pressure. If you pour mercury to depth y then the mercury will exert pressure rho g y = 13,600 Kg/m^3 * 9.8 m/s^2 * y = 133,000 N / m^3 * y. Thus the pressure in the tube will thus be atmospheric pressure + mercury pressure = 101,000 N/m^2 + 133,000 N/m^3 * y. As a result the altitude of the air column will be altitude of air column when y cm of mercury are supported: altitude of air column = atmospheric pressure / (atmospheric pressure + mercury pressure) * .9 m =101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m. At the point where mercury spills over the altitude of the air column will be .9 m - y. Thus at this point 101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m = .9 m - y. This equation can be solved for y. The result is y = .14 m, approx. The pressure will be 101,000 N/m^2 + 133,000 N/m^3 * .14 m = 120,000 N/m^2. The gauge pressure will therefore be 120,000 N/m^2 - 101,000 N/m^2 = 19,000 N/m^2. **
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RESPONSE --> This is very involved and confusing. There is alot assumed.
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18:19:07 query univ phy 18.75 (16.61 10th edition) univ phy problem 16.61 for what mass is rms vel .1 m/s; if ice how many molecules; if ice sphere what is diameter; is it visible? Give your solution to the problem.
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RESPONSE --> What does rms stand for? vrms = sqrt(3kt /m) I cannot find this problem but I found one similar to it as 18.75 but it says nothing about ice.
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18:21:58 ** We can solve this problem knowing that ave KE per particle is 3/2 k T so the .5 m v^2 = 3/2 k T, where v is RMS velocity. Thus m = 3 k T / v^2. Then from the density of water and the mass of the particle we can determine its volume, which is equal to 4/3 pi r^3. From this we find r. We obtain volume 3 k T / (v^2 rho), where rho is the density of water. Setting this equal to 4/3 pi r^3 we get r = [ 9 k T / ( 4 v^2 rho) ] ^(1/3). From the mass, Avogadro's number and the mass of a mole of water we determine the number of molecules. The following analysis shows the intermediate quantities we obtain in the process. Some of the calculations, which were done mentally, might be in error so you should redo them using precise values of the constants. At 273 Kelvin we have ave KE = 3/2 k T = 5.5 * 10^-21 Joules. mass is found by solving .5 m v^2 = 3/2 k T for m, obtaining m = 3/2 k T / (.5 v^2) = 5.5 * 10^-21 J / (.5 * (.001 m/s)^2 ) = 1.2 * 10^-14 kg. The volume of the sphere is therefore 1.2 * 10^-14 kg / (1000 kg / m^2) = 1.2 * 10^-17 m^3. Setting this equal to 4/3 pi r^3 we obtain radius r = ( 1.2 * 10^-17 m^3 / 4.2)^(1/3) = ( 2.8 * 10^-18 m^3)^(1/3) = 1.4 * 10^-6 m. Diameter is double this, about 2.8 * 10^-6 m. This is only 3 microns, and is not visible. A water molecule contains 2 hydrogen and 1 oxygen molecule with total molar mass 18 grams = .018 kg. The 1.2 * 10^-14 kg mass of particle therefore consists of 1.2 * 10^-14 / (.018 kg / mole) = 6.7 * 10^-13 moles. With about 6 * 10^23 particles in a mole this consists of 6.7 * 10^-13 moles * 6 * 10^23 particles / mole = 4* 10^11 particles (about 400 billion water molecules). **
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RESPONSE --> I didn't have any of this. To find the velocity of molecule average KE is .5 m v^2 = 3/2 k T.
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pŘγ]yP깞O LxŐ assignment #014 014. `query 4 Physics II 06-29-2008
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20:22:17 query introset plug of water from cylinder given gauge pressure, c-s hole area, length of plug. Explain how we can determine the velocity of exiting water, given the pressure difference between inside and outside, by considering a plug of known cross-sectional area and length?
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RESPONSE --> If the pressure inside and outside is different, then the gauge pressure tells how many N/ m^2 times the c-s area will get Force in N. The work done is the force time the length of the plug. This work is equal to the KEfinal = .5 m v^2 solve for v.
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20:22:59 ** The net force on the plug is P * A, where A is its cross-sectional area and P the pressure difference. If L is the length of the plug then the net force P * A acts thru dist L doing work P * A * L. If init vel is 0 and there are no dissipative forces, this is the kinetic energy attained by the plug. The volume of the plug is A * L so its mass is rho * A * L. Thus we have mass rho * A * L with KE equal to P * A * L. Setting .5 m v^2 = KE we have .5 rho A L v^2 = P A L so that v = sqrt( 2 P / rho). **
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20:24:51 prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.
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RESPONSE --> This can be solved byfinding the volume given = 51.072 m^3 d = m/v m= d*v if the density of the air is known solve for mass
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20:25:31 The density of air at common atmospheric temperature and pressure it about 1.3 kg / m^3. The volume of the room is 4.8 m * 3.8 m * 2.8 m = 51 m^3, approximately. The mass of the air in this room is therefore mass = density * volume = 1.3 kg / m^3 * 51 m^3 = 66 kg, approximately. This is a medium-sized room, and the mass is close to the average mass of a medium-sized person.
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RESPONSE -->
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20:27:06 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
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RESPONSE --> assuming i can skip since in university physics like in assignment 11
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20:28:02 The density of blood is close to that of water, 1000 kg / m^3. So the pressure difference corresponding to a height difference of 1.6 m is pressure difference = rho g h = 1000 kg/m^3 * 9.80 m/s^2 * 1.60 m = 15600 kg / ( m s^2) = 15600 (kg m / s^2) / m^2 = 15600 N / m^2, or 15600 Pascals. 1 mm of mercury is 133 Pascals, so 15,600 Pa = 15,600 Pa * ( 1 mm of Hg / 133 Pa) = 117 mm of mercury. Blood pressures are measured in mm of mercury; e.g. a blood pressure of 120 / 70 stands for a systolic pressure of 120 mm of mercury and a diastolic pressure of 70 mm of mercury.
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RESPONSE -->
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21:00:24 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
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RESPONSE --> The volume of the balloon is 4/3 pi r^3. V= 1663 m^3 The buoyant force is the force keeping the ballon in the air so it is equal and opposite gravity force which is m*g = 9123 N = buoyant force.
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21:05:09 ** The volume of the balloon is about 4/3 pi r^3 = 1660 cubic meters and mass of air displaced is about 1.3 kg / m^3 * 1660 m^3 = 2160 kg. The buoyant force is equal in magnitude to the force of gravity on the displaced air, or about 2160 kg * 9.8 m/s^2 = 20500 Newtons, approx.. If the total mass of the balloon, including helium, is 930 kg then the net force is about buoyant force - weight = 20,500 N - 9100 N = 11,400 N If the 930 kg doesn't include the helium we have to account also for the force of gravity on its mass. At about .18 kg/m^3 the 1660 m^3 of helium will have mass about 300 kg on which gravity exerts an approximate force of 2900 N, so the net force on the balloon would be around 11,400 N - 2900 N = 8500 N approx. The mass that can be supported by this force is m = F / g = 8500 N / (9.8 m/s^2) = 870 kg, approx.. **
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RESPONSE --> The air in the balloon also has mass but the balloon's apparent weight is only 9 30 kg so the bouyant force must be counteracting the gravity.
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21:14:27 univ 14.51 (14.55 10th edition) U tube with Hg, 15 cm water added, pressure at interface, vert separation of top of water and top of Hg where exposed to atmosphere. Give your solution to this problem.
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RESPONSE --> the pressure between the water and the mercury is due to the water. potential pressure = d g h = 1000* 9.81* .15 = 1471.5 Pa as the gauge pressure. I couldn't figure out how to calculate the height on the other side. The difference in pressures on each side could be how.
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21:19:06 ** At the interface the pressure is that of the atmosphere plus 15 cm of water, or 1 atm + 1000 kg/m^3 * 9.8 m/s^2 * .15 m = 1 atm + 1470 Pa. The 15 cm of water above the water-mercury interface must be balanced by the mercury above this level on the other side. Since mercury is 13.6 times denser than water there must therefore be 15 cm / (13.6) = 1.1 cm of mercury above this level on the other side. This leaves the top of the water column at 15 cm - 1.1 cm = 13.9 cm above the mercury-air interface. } Here's an alternative solution using Bernoulli's equation, somewhat more rigorous and giving a broader context to the solution: Comparing the interface between mercury and atmosphere with the interface between mercury and water we see that the pressure difference is 1470 Pa. Since velocity is zero we have P1 + rho g y1 = P2 + rho g y2, or rho g (y1 - y2) = P2 - P1 = 1470 Pa. Thus altitude difference between these two points is y1 - y2 = 1470 Pa / (rho g) = 1470 Pa / (13600 kg/m^2 * 9.8 m/s^2) = .011 m approx. or about 1.1 cm. The top of the mercury column exposed to air is thus 1.1 cm higher than the water-mercury interface. Since there are 15 cm of water above this interface the top of the water column is 15 cm - 1.1 cm = 13.9 cm higher than the top of the mercury column. NOTE BRIEF SOLN BY STUDENT: Using Bernoullis Equation we come to: 'rho*g*y1='rho*g*y2 1*10^3*9.8*.15 =13.6*10^3*9.8*y2 y2=.011 m h=y1-y2 h=.15-.011=.139m h=13.9cm. **
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RESPONSE --> The difference in density of the two chemicals means that the one with the lower density mercury will end up changing its height less then the water would. So rho g (y1 - y2) = P2 - P1 = 1470 Pa the difference in height is then 1.1cm which makes the water 13.9 cm higher.
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