course Phy 201 ҠYkRassignment #021
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14:14:04 `q001. Note that this assignment contains 3 questions. . A projectile has an initial velocity of 12 meters/second, entirely in the horizontal direction. After falling to a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's velocity vector?
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RESPONSE --> Well, the horizontal velocity when the object hits the floor will remain 12 m/s. As for the horizontal velocity, it will be vo=0, a=9.8cm/s^2, and 'ds=3m vf^2=vo^2+2a'ds =0+2(9.8)*3 =58.8 m/s =sqrt (58.8 m/s) =7.7 m/s Since 12 m/s is in the horizontal direction and 7.7 is in the vertical, we will take the first as the x and the second as the y and can fill them into the PT. 12^2+7.7^2=c^2 144+59.29=c^2 203.29=c^2 14.3=c - This is the magnitude The direction of the magnitude can be found by using arctan (y/x) Since 7.7 is on the negative y axis, it will be negative, whereas the x will remain positive since it is on the positive side of the x axis. arctan(-7.7/12)=-32.7 Since y is negative, we add 360 to -32.7, yielding a direction of magnitude of about 327 degrees. confidence assessment: 3
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14:14:19 To answer this question we must first determine the horizontal and vertical velocities of the projectile at the instant it first encounters the floor. The horizontal velocity will remain at 12 meters/second. The vertical velocity will be the velocity attained by a falling object which is released from rest and allowed to fall three meters under the influence of gravity. Thus the vertical motion will be characterized by initial velocity v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8 meters/second ^ 2. The fourth equation of motion, vf^2 = v0^2 + 2 a `ds, yields final vel in y direction: vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second. Since we took the acceleration to be in the positive direction the final velocity will be + 7.7 meters/second. This final velocity is in the downward direction. On a standard x-y coordinate system, this velocity will be directed along the negative y axis and the final velocity will have y coordinate -7.7 m/s and x coordinate 12 meters/second. The magnitude of the final velocity is therefore `sqrt((12 meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second, approximately. The direction of the final velocity will therefore be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees, very approximately, as measured in the counterclockwise direction from the positive x axis. The direction of the projectile at this instant is therefore 35 degrees below horizontal. This angle is more commonly expressed as 360 degrees + (-35 degrees) = 325 degrees.
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RESPONSE --> I understand. self critique assessment: 3
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й_M_ assignment #022 022. Motion in force field Physics II 06-19-2007 ᧖o}yPن assignment #023 023. Forces (atwood, chains) Physics II 06-19-2007
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14:35:06 `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
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RESPONSE --> a=9.8m/s^2 'ds=12 m v0=20 m/s We can figure out the initial velocity in the vertical direction by using the initial velocity y and the formula we know that can be used to determine velocity for x and y components. y=vsin(theta) y=20sin(30)=10 m/s So, since we filled in initial velocity as the coefficient, 10 m/s represents the initial velocity of the object as it travels the vertical direction. So now we know 'ds=-12 m (because the object is traveling downward), the acceleration is -9.8 m/s^2 (again, b/c its force is downward), and now vertical vo=10m/s. vf^2=10^2+2(-9.8)(-12) =100+235.2 =335.2 =sqrt(335) =-18 m/s vf+vo/2=vAve =-18+10/2 =-8/2 =-4=vAve vAve='ds/'dt -4=-12/'dt -4*'dt=-12 'dt=3seconds confidence assessment: 2
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14:35:46 To determine the time required to reach the level surface we need only analyze the vertical motion of the projectile. The acceleration in the vertical direction will be 9.8 meters/second ^ 2 in the downward direction, and the displacement will be 12 meters in the downward direction. Taking the initial velocity to be upward into the right, we situate our x-y coordinate system with the y direction vertically upward and the x direction toward the right. Thus the initial velocity in the vertical direction will be equal to the y component of the initial velocity, which is v0y = 20 meters/second * sine (30 degrees) = 10 meters/second. Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8 meters/second ^ 2, we see that we can find the time `dt required to reach the level surface using either the third equation of motion `ds = v0 `dt + .5 a `dt^2, or we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find vf after which we can easily find `dt. To avoid having to solve a quadratic in `dt we choose to start with the fourth equation. We obtain vf = +-`sqrt ( (10 meters/second) ^ 2 + 2 * (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second, approximately. Since we know that the final velocity will be in the downward direction, we choose vf = -18.3 meters/second. We can now find the average velocity in the y direction. Averaging the initial 10 meters/second with the final -18.3 meters/second, we see that the average vertical velocity is -4.2 meters/second. Thus the time required for the -12 meters displacement is `dt = `ds / vAve = -12 meters/(-4.2 meters/second) = 2.7 seconds.
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RESPONSE --> I understand. Sometimes the ""directions"" get hard for me to visualize and conceptualize, but it seems as though if I don't overthink it, I can apply the ideas well enough. self critique assessment: 3
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14:39:14 `q003. What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes the flat surface.
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RESPONSE --> It seems that we can just do for x now what we already did for y--find the velocity. If we can find that velocity, we know the time, and from the equation vAve='ds/'dt we see that multiplying the velocity by the time gives the displacement. So Vx=20m/s*cos(30)=17.3 m/s 17.3 m/s * 3 seconds=51.9 meters confidence assessment: 3
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14:39:23 The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing that the projectile travels for 2.7 seconds we can easily find the horizontal range. The horizontal velocity of the projectile is simply the x component of the velocity: horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second. Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters, approximately.
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RESPONSE --> I understand self critique assessment: 3
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15:49:12 `q001. Note that this assignment contains 2 questions, which relate to a force-field experiment which is done using a computer simulation, and could for example represent the force on a spacecraft, where uphill and downhill are not relevant concepts. . An object with a mass of 4 kg is traveling in the x direction at 10 meters/second when it enters a region where it experiences a constant net force of 5 Newtons directed at 210 degrees, as measured in the counterclockwise direction from the positive x axis. How long will take before the velocity in the x direction decreases to 0? What will be the y velocity of the object at this instant?
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RESPONSE --> Since the acceleration in this experiment can't be assumed to be the constant 9.8cm/s^2, we can use the information given to first find the acceleration. a=fnet/m =5N/4kg =1.25m/s^2 x=1.25*cos(210)=-1.08m/s^2 y=1.25*sin(210)=-.625m/s^2 aAve='dv/'dt We, for x and y, know aAve and 'dv. From this information, we can figure out 'dt. For horizontal 'dt 'dt='dv/aAve =-10m/s/-1.08m/s^2 =9.2 seconds For y, we know aAve and 'dt, which we can use to find 'dv. -.625*9.2 seconds=-5.72 m/s, which will be the velocity of the object as it reaches zero. confidence assessment: 2
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15:49:47 A constant net force of 5 Newtons on a 4 kg object will result in an acceleration of 5 Newtons/(4 kg) = 1.25 meters/second ^ 2. If the force is directed at 210 degrees then the acceleration will also be directed at 210 degrees, so that the acceleration has x component 1.25 meters/second ^ 2 * cosine (210 degrees) = -1.08 meters/second ^ 2, and a y component of 1.25 meters/second ^ 2 * sine (210 degrees) = -.63 meters/second ^ 2. We analyze the x motion first. The initial velocity in the x direction is given as 10 meters/second, we just found that the acceleration in the x direction is -1.08 meters/second ^ 2, and since we are trying to find the time required for the object to come to rest the final velocity will be zero. We easily see that the change in the next velocity is -10 meters/second. At a rate of negative -1.08 meters/second ^ 2, the time required for the -10 meters/second change in velocity is `dt = -10 meters/second / (-1.08 meters/second ^ 2) = 9.2 seconds. We next analyze the y motion. The initial velocity in the y direction is zero, since the object was initially moving solely in the x direction. The acceleration in the y direction is -.63 meters/second ^ 2. Therefore during the time interval `dt = 9.2 seconds, the y velocity changed by (-.63 meters/second ^ t) * (9.2 seconds) = -6 meters/second, approximately. Thus the y velocity changes from zero to -6 meters/second during the 9.2 seconds required for the x velocity to reach zero.
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RESPONSE --> I understand self critique assessment: 3
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16:55:52 `q002. Suppose that the mass in the preceding problem encounters a region in which the force was identical to that of the problem, but that this region extended for only 30 meters in the x direction (assume that there is the limit to the extent of the field in the y direction). What will be the magnitude and direction of the velocity of the mass as it exits this region?
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RESPONSE --> x: 'ds=30m vo=10m/s acceleration=-1.08 m/s^2 vf^2=vo^2+2a'ds =100+2(-1.08)*30 =sqrt (35.2) m/s =about 6 m/s 6+10/2=8 m/s vAve 'ds/vAve='dt 30m/8m/s=3.75s Now for y, we know the acceleration (from the previous question) is -.63 m/s^2. The time, as found before, is 3.75 seconds. If we multiply the acceleration by the time, I can get the average change in velocity. So, 3.75s*-.63m/s^2=-2.4m/s (about) So, vAve's x=6 m/s and y=-2.4 m/s We use these values to find the magnitude and the direction of the magnitude. 6^2+(-2.4)^2=c^2 36+5.75=c^2 41.75=c^2 6.46=c <--magnitude arctan(-2.4/6)=-21.8+360 (bc y is negative)=338.2 degrees <--direction of magnitude confidence assessment: 3
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16:56:02 As we have seen in the preceding problem the object will have an acceleration of -1.08 meters/second ^ 2 in the x direction. Its initial x velocity is 10 meters/second and it will travel 30 meters in the x direction before exiting the region. Thus we have v0, a and `ds, so that you to the third or fourth equation of uniform accelerated motion will give us information. The fourth equation tells us that vf = +-`sqrt( (10 meters/second) ^ 2 + 2 * (-1.08 meters/second ^ 2) * (30 meters) ) = +-6 meters/second. Since we must exit the region in the positive x direction, we choose vf = + 6 meters/second. It follows that the average x velocity is the average of the initial 10 meters/second and the final 6 meters/second, or eight meters/second. Thus the time required to pass-through the region is 30 meters/(8 meters/second) = 3.75 seconds. During this time the y velocity is changing at -.63 meters/second ^ 2. Thus the change in the y velocity is (-.63 meters/second ^ 2) * (3.75 seconds) = -2.4 meters/second, approximately. Since the initial y velocity was zero, the y velocity upon exiting the region will be -2.4 meters/second. Thus when exiting the region the object has velocity components +6 meters/second in the x direction and -2.4 meters/second in the y direction. Its velocity therefore has magnitude `sqrt ( (6 meters/second) ^ 2 + (-2.4 meters/second) ^ 2) = 6.4 meters/second. The direction of velocity will be arctan ( (-2.4 meters/second) / (6 meters/second) ) = -22 degrees, approximately. Thus the object exits at 6.4 meters/second at an angle of 22 degrees below the positive x axis, or at angle -22 degrees + 360 degrees = 338 degrees.
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RESPONSE --> I understand. self critique assessment: 3
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20:04:06 `q001. Note that this assignment contains 3 questions. . A chain 200 cm long has a density of 15 g/cm. Part of the chain lies on a tabletop, with which it has a coefficient of friction equal to .10. The other part of the chain hangs over the edge of the tabletop. If 50 cm of chain hang over the edge of the tabletop, what will be the acceleration of the chain?
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RESPONSE --> 50 cm of chain is hanging over the edge, and for each centimeter of length the chain has a mass of 15 g. So, the portion of the chain hanging over the edge has mass 50*15=750g (or .75 kg). To find the force of gravity acting on the chain that hangs over the edge, we will multiply the mass by the acceleration. .75kg*9.8m/s^2=7.35N The rest of the chain has length of 150 cm. 150 cm * 15 kg/cm=2250g or 2.25 kg 2.25kg*9.8m/s^2=22.05N We were told that this force has a .1 friction acting against it, meaning that 22.05*.1=2.2N frictional force So, the hanging force onthe change is 7.35N and the friction force is 2.2 N. Obviously, the friction force is acting opposite the hanging force and hence will be negative whereas the hanging force of gravity is positve. 7.35N-2.2N=5.15N (net force) Now we know the net force of the chain, so all we need to find is the total mass in order to be able to calculate the acceleration of the whole chain. 200cm of chain * 15kg/cm=3000g=3kg for the total mass a=Fnet/m =5.15N/3kg=1.72 cm/s^2 confidence assessment: 3
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20:04:13 The part of the chain hanging over the edge of the table will experience an unbalanced force from gravity and will therefore tend to accelerate chain in the direction of the hanging portion. The remainder of the chain will also experience the gravitational force, but this force will be countered by the upward force exerted on the chain by the table. The force between the chain and the table will give rise to a frictional force which will resist motion toward the hanging portion of the chain. If 50 cm of chain hang over the edge of the tabletop, then we have 50 cm * (15 g/cm) = 750 grams = .75 kg of chain hanging over the edge. Gravity will exert a force of 9.8 meters/second ^ 2 * .75 kg = 7.3 Newtons of force on this mass, and this force will tend to accelerate the chain. The remaining 150 cm of chain lie on the tabletop. This portion of the chain has a mass which is easily shown to be 2.25 kg, so gravity exerts a force of approximately 21 Newtons on this portion of the chain. The tabletop pushes backup with a 21 Newton force, and this force between tabletop and chain results in a frictional force of .10 * 21 Newtons = 2.1 Newtons. We thus have the 7.3 Newton gravitational force on the hanging portion of the chain, resisted by the 2.1 Newton force of friction to give is a net force of 5.2 Newtons. Since the chain has a total mass of 3 kg, this net force results in an acceleration of 5.2 N / (3 kg) = 1.7 meters/second ^ 2, approximately.
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RESPONSE --> I understand. self critique assessment: 3
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20:05:07 `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
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RESPONSE --> I am really not sure how to approach this one... confidence assessment: 0
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20:18:37 The maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain. If x stands for the length in cm of the portion of chain hanging over the edge of the table, then the mass of the length is x * .015 kg / cm and it experiences a gravitational force of (x * .015 kg / cm) * 9.8 m/s^2 = x * .147 N / cm. The portion of chain remaining on the tabletop is 200 cm - x. The mass of this portion is (200 cm - x) * .015 kg / cm and gravity exerts a force of (200 cm - x) * .015 kg / cm * 9.8 meters/second ^ 2 = .147 N / cm * (200 cm - x) on this portion. This will result in a frictional force of .10 * .147 N / cm * (200 cm - x) = .0147 N / cm * (200 cm - x). Since the maximum length that can hang over is the length for which the frictional force opposing motion is precisely equal to the gravitational force on the hanging portion of the chain, we set the to forces equal and solve for x. Our equation is .0147 N / cm * (200 cm - x) = .147 N/cm * x. Dividing both sides by .0147 N/cm we obtain 200 cm - x = 10 * x. Adding x to both sides we obtain 200 cm = 11 x so that x = 200 cm / 11 = 18 cm, approx..
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RESPONSE --> I will explain this in my own words: The frictional force has to equal the gravitational force for the chain to not accelerate. Because we have to find the mass of the chain by multiplying the centimeters by the kg/cm but we dont know the length in cm, x represents length but 15kg/cm is constant, as is the acceleration by which the mass is divided. Since there is still a variable, though, the equation remains x*.015kg/9.8cm/s^2. We can go ahead and multiply through the last two numbers, leaving the variable x*.147N/cm. Since we know the wholechain is 200cm in length, we know that 200-x (the unknown length required to keep the chain balanced and not accelerating). We treat this portion of the equation the same way we did the other, except that the forces of both gravity and friction are acting on this part of the chain. So, 200-x*.015kg/cm*9.8cm/s^2=.147N/cm*(200 cm-x). The frictional force is .1 of the gravitational newton force, so .1* .147N/cm*(200 cm-x)=.0147N/cm*200cm-x Since the forces have to be equal, we can set the two equations with variables equal to each other, since they each represent the forces exerted on either part of the chain. x*.147N/cm=.0147N/cm*200cm-x We follow the order of operations until we can solve directly for x. Then, we do so, yielding about 18 cm of length that can hang over for the system to not be accelerating. self critique assessment: 2
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20:25:33 `q003. The air resistance encountered by a certain falling object of mass 5 kg is given in Newtons by the formula F = .125 v^2, where the force F is in Newtons when the velocity v is in meters/second. As the object falls its velocity increases, and keeps increasing as it approaches its terminal velocity at which the net force on the falling object is zero, which by Newton's Second Law results in zero acceleration and hence in constant velocity. What is the terminal velocity of this object?
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RESPONSE --> The object will encounter gravity, which will push downward, and air resistance, which will work to counter gravity. We know the object is mass 5kg, so we can multiply this by the acceleration of gravity and get the total force of gravity on the object: 5kg*9.8cm/s^2=49N Now, we literally use the equation given. We fill in 49 N for the force and then solve for v^2. 49N=.125v^2 392=v^2 19.8m/s=v confidence assessment: 3
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20:25:40 Only two forces act on this object, the downward force exerted on it by gravity and the upward force exerted by air resistance. The downward force exerted by gravity remains constant at 5 kg * 9.8 meters/second ^ 2 = 49 Newtons. When this force is equal to the .125 v^2 Newton force of friction the object will be at terminal velocity. Setting .125 v^2 Newtons = 49 Newtons, we divide both sides by .125 Newtons to obtain v^2 = 49 Newtons/(.125 Newtons) = 392. Taking square roots we obtain v = `sqrt (392) = 19.8, which represents 19.8 meters/second.
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RESPONSE --> I understand. self critique assessment: 3
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