course Phy 121 B稜[ӡHxwJƎFStudent Name:
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23:08:34 `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 --> Total L = 200cm d = 15g/cm f coefficient = 0.10 L over edge = 50cm We need to know the mass of the part of the chain that is over the edge. We multiply the length of that section by the density of the entire chain.(50cm)(15g/cm) = 750g = 0.75kg The mass of the rest of the chain on the table is found in the same manner. (150cm)(15g/cm) = 2250g = 2.25kg Force hanging = (0.75kg)(9.8m/s^2) = 7.35 N Force on table = (2.25kg)(9.8m/s^2) = 22.05 N The frictional force would only apply to the part of the chain on the table because there is no friction on a hanging object. Therefore, the force of friction would be 0.10 or 10% of the force on the section on the table. f = 0.10 * 22.05N = 2.2 N Now we can find net force of the hanging section by combining the gravitational force with the frictional force on the table which works against it. We know the total mass of the chain is 200cm * 15g/cm = 3000g = 3kg. With these two elements, we can calculate the acceleration. Fnet = 7.35N - 2.2N = 5.15N a = Fnet/m = 5.15N/3kg = 1.72 m/s^2
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23:09:18 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.
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23:13:18 `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
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RESPONSE --> In the previous problem we used the length, density, mass, and acceleration of gravity to find the force. Now we can use the acceleration of gravity and density to find the length. Total length = 200cm = 2m g = 9.8m/s^2 d = 15g/cm = 1.5kg/m hanging F = (L * d)(g) = x(1.5kg/m)(9.8m/s^2) = x(14.7kg/s^2) m = L*d = (2m - x)(1.5kg/m) = 3kg - x(1.5kg/m) F = m*g = [3kg - x(1.5kg/m)] * 9.8m/s^2 = 29.4N - x(14.7kg/s^2)
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23:14:45 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 --> Why is my answer greater than yours? Which step did I do wrong?
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23:19:32 `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 --> falling F = (5kg)(9.8m/s^2) = 49N f = opposing force = 0.125 v^2 N When velocity is constant, the opposing forces will be equal. Therefore, we can set the gravitational force and force of air resistance equal to each other and solve for the velocity. 49N = 0.125N * v^2 v^2 = 392N v = 19.80 m/s
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23:24:09 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 thought this problem was fairly simple.
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ٷ᷅_녔гZ assignment #023 023. `query 23 Physics I 07-06-2006
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17:52:43 Query gen phy 7.27 bumper cars 450 kg at 4.5 m/s, 550 kg at 3.7 m/s, collision from back, elastic
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RESPONSE --> m1 = 450kg v1 = 4.5m/s m2 = 550kg v2 = 3.7m/s Velocities after collision: Since the collision is elastic, we know that the mass and volume relationship can be expressed by the formula, m1v1(before) + m2v2(before) = m1v1(after) + m2v2(after). The relationship for the velocities before and after the collision is expressed by the formula, v2 - v1 = v1(after) - v2(after) We can use the elements we know and use substitution to find the mass and velocity after the collision: 450kg*4.5m/s + 550kg*3.7m/s = 450kg*v1 + 550kg*v2 4060kg*m/s = 450kg*v1 + 550kg*v2 Kilograms can be factored out of the equation: 4060m/s = 450*v1 + 550*v2 Also, we can use the velocities before collision to find the equivalent of the velocities after collision. 3.7m/s - 4.5m/s = v1(after) - v2(after) -0.8m/s = v1 - v2 v2 = 0.8m/s + v1 4060m/s = 450*v1 + 550(0.8m/s + v1) 4060m/s = 450*v1 + 440m/s + 550*v1 3620m/s = 1000*v1 v1 = 3.62m/s v2 = 0.8m/s + 3.62m/s = 4.42m/s Change in momentum of each: momentum = m*`dv `dp1 = m1*`dv1(after) - m1`dv1(before) = 450kg*3.62m/s - 450kg*4.5m/s = -396 kg*m/s `dp2 = m1*`dv1(after) - m1`dv1(before) = 550kg*4.42m/s - 550kg*3.7m/s = 396 kg*m/s
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17:54:44 ** For an elastic collision we have m1 v1 + m2 v2 = m1 v1' + m2 v2' and v2 - v1 = -( v2' - v1'). We substitute m1, v1, m2 and v2 to obtain 450 kg * 4.5 m/s + 550 kg * 3.7 m/s = 450 kg * v1 ' + 550 kg * v2 ', or 4060 kg m/s = 450 kg * v1 ' + 550 kg * v2 ' . Dividing by 10 and by kg we have 406 m/s = 45 v1 ' + 55 v2 '. We also obtain 3.7 m/s - 4.5 m/s = -(v2 ' - v1 ' ) or v1 ' = v2 ' - .8 m/s. Substituting this into 406 m/s = 45 v1 ' + 55 v2 ' we obtain 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' . We easily solve for v2 ' obtaining v2 ' = 4.42 m/s. This gives us v1 ' = 4.42 m/s - .8 m/s = 3.62 m/s. Checking to be sure that momentum is conserved we see that the after-collision momentum is pAfter = 450 kg * 3.62 m/s + 550 kg * 4.42 m/s = 4060 m/s. The momentum change of the first car is m1 v1 ' - m1 v1 = 450 kg * 3.62 m/s - 450 kg * 4.5 m/s = - 396 kg m/s. The momentum change of the second car is m2 v2 ' - m2 v2 = 550 kg * 4.42 m/s - 550 kg * 3.7 m/s = + 396 kg m/s. Momentum changes are equal and opposite. NOTE ON SOLVING 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' FOR v2 ': Starting with 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' use the Distributive Law to get 406 m/s = 45 v2 ' - 36 m/s + 55 v2 ' then collect the v2 ' terms to get 406 m/s = -36 m/s + 100 v2 '. Add 36 m/s to both sides: 442 m/s = 100 v2 ' so that v2 ' = 442 m/s / 100 = 4.42 m/s. *
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RESPONSE --> I thought I'd try this one even though it's not for my course.
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17:55:48 Univ. 3.48. (not in 11th edition) ball 60 deg wall 18 m away strikes 8 m higher than thrown. What are the Init speed of the ball and the magnitude and angle of the velocity at impact?
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RESPONSE --> I am not required to answer questions for University Physics.
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17:56:24 ** We know the following: For y motion `dsy = + 8 m, ay = -g = - 9.8 m/s^2 and v0y = v0 sin(60 deg) = .5 v0. For x motion `dsx = 18 m, ax = 0 and v0x = v0 cos(60 deg) = .87 v0, approx. Assuming a coordinate system where motion starts at the origin: The equation of motion in the x direction is thus x = .5 v0 * t and the equation of y motion is y = .87 v0 t - .5 g t^2. We know x and y at impact and we know g so we could solve these two equations simultaneously for v0 and t. We begin by eliminating t from the two equations: x = .5 v0 * t so t = 2 x / v0. Substituting this expression for t in the second equation we obtain y = .87 v0 * (2 x / v0) - .5 g ( 2 x / v0) ^ 2. Multiplying both sides by v0^2 we obtain v0^2 y = .87 v0^2 ( 2 x) - .5 g * 4 x^2. Bringing all the v0 terms to the left-hand side we have v0^2 y - 1.73 v0^2 x = -2 g * x^2. Factoring v0 we have v0^2 ( y - 1.73 x) = -2 g x^2 so that v0 = +-sqrt(-2 g x^2 / ( y - 1.73 x) ) ) . Since we know that y = 8 m when x = 18 m we obtain = +- sqrt( -9.8 m/s^2 * (18 m)^2 / ( 8 m - 1.73 * 18 m) ) = +-sqrt(277 m^2 / s^2) = +-16.7 m/s, approx.. We choose the positive value of v0, since the negative value would have the projectile moving 'backward' from its starting point. Substituting this value into t = 2 x / v0 and recalling that our solution applies to the instant of impact when x = 16 m and y = 8 m we obtain t = 2 * 18 m / (16.7 m/s) = 2.16 s. Alternatively we can solve the system for v0 less symbolically and perhaps gain different insight into the meaning of the solution. Starting with the equations x = .5 v0 * t and y = .87 v0 t - .5 g t^2 we see that impact occurs when x = .5 v0 t = 18 m so that t = 18 m / (.5 v0) = 36 m / v0. At this instant of impact y = 8 m so substituting this and the t just obtained into the equation of motion for y we get y = .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m. The equation .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m is easily solved for v0, obtaining v0 = 16.7 m/s. With this initial velocity we again confirm that t = 2.16 sec at impact. Note that at t = 2.16 sec we get y = 14.4 m/s * 2.16 s - 4.9 m/s^2 * (2.16 s)^2 = 8 m, within roundoff error, confirming this solution. We need the magnitude and direction of the velocity at impact. We therefore need the x and y components of the velocity at the t = 2.16 sec instant. At this instant we have x and y velocities vx = dx/dt = .5 v0 = 8.35 m/s and vy = dy/dt = 14.4 m/s - 9.8 m/s^2 * t = 14.4 m/s - 9.8 m/s^2 * 2.16 s = -5.6 m/s, approx. The velocity at impact therefore has magnitude sqrt( (8.25 m/s)^2 + (-5.6 m/s)^2 ) = 10 m/s, approx. and the angle is arctan(vy/vx) = arctan(-5.6 / 8.25) = -34 deg, approx. At impact the object is moving at 10 m/s and at 34 deg below horizontal (i.e., it's on its way back down). **
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RESPONSE --> ok
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