#$&* course PHY 202 3/1 about 11:15 am 005. `query 5
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Given Solution: ** 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 F_net = P * A acts thru distance L doing work `dW = F_net * L = P * A * L. If the initial velocity of the plug 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). STUDENT SOLUTION AND QUESTION From looking at my notes from the Intro Problem Set, it was a lot of steps to determine the velocity in that situation. The Force was determined first by using F = (P * cross-sectional area). With that obtained Force, use the work formula ‘dW = F * ‘ds using the given length as the ‘ds. Then we had to obtain the Volume of the ‘plug’ by cross-sectional area * length. That Volume will then be using to determine the mass by using m = V * d, and the density is a given as 1000kg/m^3. With that mass, use the KE equation 1/2(m * v^2) to solve for v using the previously obtained ‘dW for KE since ‘dW = ‘dKE. Some of my symbols are different than yours. I’m in your PHY 201 class right now and am in the work and energy sections. So is it wrong to use 1/2(m * v^2) instead of 1/2(‘rho * A * L * v^2)?? Aren’t they basically the same thing?? INSTRUCTOR RESPONSE You explained the process very well, though you did miss a step. m = rho * V (much better to use rho than d, which isn't really a good letter to use for density or distance, given its use to represent the prefix 'change in'). You covered this in your explanation. V isn't a given quantity; it's equal to the length of the plug multiplied by its cross-sectional area and should be expressed as such. You didn't cover this in your explanation. However, other than this one missing step, your explanation did cover the entire process. With that one additional step, your solution would be a good one. In any case, V would be A * L, and m would be rho * V = rho * A * L. So 1/2 m v^2 is the same as 1/2 rho A L v^2. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Density of air from the book is 1.29 kg/m^3 Volume of room 4.8 m * 3.8 m * 2.8 m = 51.072 m^3 Mass = ‘rho*volume = 1.29 kg/m^3 * 51.072 m^3 = 65.88 kg or about 66 kg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 of the air in that room is close to the mass of an average-sized person. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: F=PA P=F/A = mg/A, m=’rho*volume, volume = Ah, so m=’rho*A*h ‘dP=(‘rho*A*h*g)/A = ‘rho*g*h ‘dP=1050 kg/m^3 * 1.60 m * 9.8 m/s^2 = 16464 kg/(m*s^2) = 16464 Pa 16464 Pa * (1 mm Hg / 133 Pa) = 123.8 mm Hg = ‘dP confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. 760 mm of mercury is 1 atmosphere, equal to 101.3 kPa, so 1 mm of mercury is 133 Pascals (101.3 kPa / (760 mm) = 133 Pa / mm), so • 15,600 Pa = 15,600 Pa * ( 1 mm of Hg / 133 Pa) = 117 mm of mercury. (alternatively 15 600 Pa * (760 mm of mercury / (101.3 kPa ) ) = 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. The density of blood is actually a bit higher than that of water; if you use a more accurate value for the density of blood you will therefore get a slightly great result. STUDENT QUESTION I had to find that conversion online because I know that 1atm is 360mmHg but I could find a conversion for atms to Pascals to make it work from what I knew, so I had to find the 133 online. ??? INSTRUCTOR RESPONSE You should know that 1 atmosphere is about 100 kPa (more accurately 101.3 kPa but you don't need to know it that accurately), and that this is equivalent to 760 mm of mercury. Using these two measures it's easy to convert from one to the other and there's no reason to look for or try to remember a conversion directly beteween mm of mercury and Pa. Specifically the conversion factors are • 101.3 kPa / (760 mm of mercury) = 133 Pa / (mm of mercury) and • (760 mm of mercury) / (101.3 kPa) = .0075 mm of mercury / Pa If you use 100 kPa for the purposes of the problems and tests in this course, as I said before it's OK. If you're ever in a situation outside this class where you need the more accurate figure, it's easy to find. STUDENT QUESTION Do you know if our text tells us this conversion?? INSTRUCTOR RESPONSE The list of equivalent quantities is in Table 10-2 in the 6th edition. This specific conversion isn't given, but the number of Pa and the number of mm of mercury in an atmosphere both are. STUDENT COMMENT The conversion of the units here is very confusing to me. The question didn’t ask for a specific unit set, so I just assumed use the one determined by the answer (kg/ms^2=N). My answer is slightly different, because I actually looked up the density of blood, which is slightly higher than water. INSTRUCTOR RESPONSE kg/m^3 * (m/s^2) * m = kg / (m * s^2), not kg * m/s^2. kg m/s^2 is N. To get kg / (m s^2) you would have to divide kg m/s^2 by m^2. That is, you divide N by m^2, obtaining N / m^2. N / m^2 is the unit of pressure, also called the Pascal. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Volume of the sphere = (4/3)*’pi*(7.35 m)^3 = 1663.22 m^3 1163 m^3 * 1.29 kg/m^3 = 2145.27 kg F = m*a = 2145 kg * 9.8 m/s^2 = 21,021 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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 weight is 930 kg * 9.8 m/s^2 = 9100 N approx., and the net force is • Net force = buoyant force - weight = 20,500 N - 9100 N = 11,400 N, approximately. 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.. ** STUDENT QUESTION I got part of the problem right. I don’t understand the volume of air displaced….. INSTRUCTOR RESPONSE The 1660 m^3 volume of the balloon takes up 1660 m^3 that would otherwise be occupied by the surrounding air. The surrounding air would be supporting the weight of the displaced air, if the balloon wasn't there displacing it. That is, the surrounding air would act to support 20500 Newtons of air. The surrounding air is no different for the fact something else is there, instead of the air displaced air. So it supports 20500 Newtons of whatever is there displacing the air it would otherwise be supporting. This supporting force of 20500 Newtons is therefore exerted on the balloon. We call this the buoyant force of the air on the balloon. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Openstax: (a) What is the mass of a deep breath of air having a volume of 2.00 L? (b) Discuss the effect taking such a breath has on your body’s volume and density. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 2.00 L * (0.001 m^3 / 1 L) = 0.002 m^3 0.002 m^3 * 1.29 kg/m^3 = 0.00258 kg = 2.58 g Since ‘rho = mass/volume, if the mass only increases 2.58 g and the volume increases by 2000 cm^3, the density will decrease. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: PV = n R T implies that n = P V / R T. P is atmospheric pressure, V is 2 liters or .002 m^3, and T is body temperature (around 310 K). n = (100 000 N/m^2 * .002 m^3) / ( (8.31 Joules / (mole K)) * 310 k) ). The result is about 1/10 of a mole. Nitrogen makes up most of our air; with its atomic number 7 its atomic mass is 14; it is diatomic so a mole has mass 28 grams. Oxygen makes up most of the rest; with its atomic number 8 and diatomic nature this increases the molar density to around 29 grams. 1/10 of a mole of air therefore has a mass of about 3 grams. Your body's density is close to that of water. Its volume increases by 2 liters with the breath, but its mass increases by only about 3 grams (much less than the 2000 grams that would be gained by an equal amount of water). So your density will decrease. A person with a volume of 100 liters weighs around 220 pounds. Increasing the volume of that person by 2 liters corresponds to an increase of 2% in volume, with negligible increase in mass. This would correspond to about a 2% decrease in density. Your Self-Critique:I solved this differently but think it is still correct. I do understand your solution. Your Self-Critique Rating:3 ********************************************* Question: Openstax: The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of 15.0 cm^2. What force does it exert to accomplish this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P = 120 mm Hg * (133 Pa / 1 mm Hg) = 15,960 Pa = 15,960 N/m^2 A = 0.15 cm^2 * (1 m^2 / 10,000 cm^2) = 0.0015 m^2 F = P*A = 15,960 N/m^2 * 0.0015 m^2 = 23.9 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A 120 mm column of Hg corresponds to pressure change `dP = rho g h = 13600 kg / m^3 * 9.8 m/s^2 * .120 m = 15000 N/m^2 or 15000 Pa. On an area of 15.0 cm^2 this pressure would exert a force of F = P * A = 15 000 N/m^2 * (15.0 cm^2) = 225 000 N/m^2 * cm^2 = 225 000 N/m^2 * (.01 m)^2 = 22.5 N. Your Self-Critique:Again, I solved this differently but I believe this is correct also. I do understand how you solved this. Your Self-Critique Rating:3 ********************************************* Question: Openstax (optional for Principles of Physics): What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are parts by mass, not volume.) Assume that this is a simple mixture having an average density equal to the weighted densities of its constituents. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ‘rho gold = 19.32 g/cm^3 ‘rho silver = 10.501 g/cm^3 ‘rho copper = 8.80 g/cm^3 (18/24)*19.32 g/cm^3 + (5/24)*(10.501 g/cm^3)*(1/24)*(8.8 g/cm^3) = approx. 17 g/cm^3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using very approximate densities (20 000 kg/m^3 for gold, 9000 kg/m^3 for silver and 8000 kg/m^3 for copper) we obtain the following sample solution: 1 part of gold occupies 1/20 the volume of 1 part of water 1 part of silver occupies 1/9 the volume of 1 part of water 1 part of copper occupies 1/8 the volume of 1 part of water. So 18 parts gold, 5 parts silver and 1 part copper will occupy the volume of 18 * 1/20 + 5 * 1/9 + 1 * 1/8 = 569 / 360 = 1.58 (approx.) part of water and will have a mass of 18 + 5 + 1 = 24 times that of 1 part of water. The density will therefore be 24 / 1.58 = 15 times that of water. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Optional Openstax: A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Cork approx. 1 inch or 2.54 cm or 0.0254 m A_cork = ‘pi * (0.0254 m)^2 = 0.000507 m^2 P = F/A = 120 N / 0.000507 m^2 = 236823 N/m^2 A_bottom = pi * (0.07 m)^2 = 0.015 m^2 F = P*A = 236823 N/m^2 * 0.015 m^2 = 3552.35 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The 120-N force exerted on the wine by the cork was spread over an area of about 3 cm^2 (this being the approximate area of a circle 2 cm in diameter), resulting in an additional fluid pressure of 120 N / (3 cm^2) = 40 N / cm^2 = 40 N / (.01 m)^2 = 400 000 N / m^2. As a result the pressure of the wine increased by the same amount. The bottom of the jug has area approximately 150 cm^2 = .015 m^2, so it experienced an additional force of .015 m^2 * 400 000 N/m^2 = 6000 N, well over 1000 pounds. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: `q001. Water pressure exerts a force of .8 Newtons, in excess of the pressure exerted by the atmosphere on the other end, on one end of a water 'plug' with cross-sectional area 3 cm^2 and length 5 cm. The 'plug' is forced out of the side of the container by the net force as it moves through its 5 cm length, starting from rest. • How much work will the net force do on the 'plug'? • What will be the KE of the 'plug' as it exits the container? • How fast will the 'plug' be moving as it leaves the container? Answer the analogous series of questions for a 'plug with cross-sectional area 1 cm^2 and length 2 cm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ‘dW_Fnet = F*L = 0.8 N * 0.05 m = 0.04 J ‘dKE_exit = KE_f - KE_0 = 0.04 J - 0 J = 0.04 J KE = .5mv^2, m = ‘rho*A*L = 1000 kg/m^3 * 0.0003 m^2 * 0.05 m = 0.015 kg 0.04 J = .5*0.015 kg * v^2 V^2 = 0.0012 m^2/s^2 V= 0.034 m/s = 3.4 cm/s ‘dW_Fnet = 0.8 N * 0.02 m = 0.016 J KE_exit = 0.016 J M = ‘rho*A*L = 1000 kg/m^3 * 0.0001 m^2 * 0.02 m = 0.002 kg 0.016 = .5*0.002 kg * v^2 V^2 = 6.4*10^-5 m^2/s^2 V=0.008 m/s = 0.8 cm/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating:OK ********************************************* Question: `q002. A closed glass jar with a half-liter capacity has a mass of 200 grams. If it is submerged in water what will be the buoyant force acting on it, and at the instant it is released from rest what will be the net force on it and its acceleration? The drag force of water on the jar is 1 N s^2 / m^2 * v^2, where v is the velocity of the jar, then at what speed will it be rising when the net force acting on it becomes zero? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The buoyant force is equal to the weight of the volume of displaced water. 0.5 L = 0.0005 m^3 M_water = 0.0005 m^3 * 1000 kg/m^3 = 0.5 kg F_b = m*g = 0.5 kg * 9.8 m/s^2 = 4.9 N M_air in jar = 0.005 m^3 * 1.29 kg/m^3 = 0.00065 kg M_jar with air = 0.2 kg + 0.00065 kg = 0.20065 kg F_g on jar = 0.20065 kg * 9.8 m/s^2 = 1.97 N F_net with no drag = 4.9 N - 1.97 N = 2.93 N A=F_net/m = 2.93 N / 0.20065 kg = 14.6 m/s^2 I can not figure out how to do the last part of this problem. I know that when F_net =0 a=0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: 2