This is a closed book examination the use of notes book cell phones or with fell
ID: 1415279 • Letter: T
Question
This is a closed book examination the use of notes book cell phones or with fellow student or other person are forbidden. surface at 12.0 m/s in the direction. It collides with a red ball of mass 1.00 kg, which is initially at rest After the collision, the green ball travelling at 4 00 m/s In the opposite direction (a) Find the final speed and direction of the red ball, (b) Find the change in the total kinetic energy of the system which results from the collision, (c) H instead, the green and red balls stuck together after collision, how mu Ch kinetic energy would be lost? A heavy uniform solid cylinder of mass M and radius R is pivoted frictionless about its axis of symmetry. A light rope is wrapped many times around the cylinder, and a block of mass M/4 is suspended from the loose end of the rope, (a) Find the acceleration of the block, (b) find the tension in the string, (c) If the system is initially at rest, use conservation of energy to find the velocity of the block after it has fallen through a vertical distance h. A flywheel of mass 10.0 kg and radius 0 20 m starts from rest and rotates about a fixed axis, accelerating with angular acceleration 2.00 rad/s', let us consider a point P on the rim of the wheel. At t = 2.00 s, find (a) the angular velocity of the wheel (b) the speed of point P (c) the centripetal acceleration of point P (d) the net torque on the wheel, assuming it is a uniform disk (moment inertia of the disk with an axis through the center is MR^3/2, where R is the radius). (a) It is desired to place a communication satellite into orbit so that it remains fixed above a given spot on the equator of the rotating earth. What's the height above the earth of such an orbit? Consider a neutron star with a mass M equal to the mass of Sun. 1.99 times 10^10 kg. and a radius R of 12.C km. What is the free fall acceleration at its surface? A wheel with moment of inertia 127 kgm} is rotating with an angular speed of 900 rev/min on a she whose moment of inertia is negligible. A second wheel, initially at rest and with moment of inertia A.J kgm1 is suddenly coupled to the same shaft. What is the final angular velocity of the results combination of the shaft and the two wheels? A driver's manual states that a driver travelling at 12.0 m/s and desiring to stop as quickly as poss. travels 10.0 m before the foot reaches the brake. The car travels an additional 21.0 m before comic rest, (a) What coefficient of friction is assumed in these calculations? (b) What is the minimum radii turning a corner at 12.0 m/s without skidding? 7. An 1100 kg car is travelling at 12.0 m/s on a level road. The brakes are applied long enough to v 51.0 kJ of kinetic energy, (a) What is the final speed of the car? (b) How mu Ch more kinetic energy be removed by the brakes to stop the car?Explanation / Answer
1) a) 0.25 * 12 = - 0.25 * 4 + 1 * v
=> v = 4 m/sec ( + x direction)
b) change in kinetic energy = 1/2 * 0.25 * 12 * 12 - 1/2 * 0.25 * 4 * 4 - 1/2 * 1 * 4 * 4
= 8 J
c) kinetic energy lost = 1/2 * 0.25 * 12 * 12 - 1/2 * 1.25 * 2.4 * 2.4
= 14.4 J
3) a) angular velocity = 2 * 2 = 4 rad/sec
b) speed at point P = 0.20 * 4 = 0.8 m/sec
c) centripetal acceleration = 0.8 * 0.8/0.2 = 3.2 m/sec2
d) net torque = 0.5 * 10 * 0.2 * 0.2 * 2 = 0.4 N.m
5) Final angular velocity = 1.27 * 900/(6.07)
= 188.303 rev/min
6) a) coefficient of friction = (12 * 12)/(2 * 21 * 9.8)
= 0.35
b) minimum radius = (12 * 12)/(0.35 * 9.8)
= 42 m
7) a) Final speed = sqrt(2 * 28200/1100)
= 7.16 m/sec
b) kinetic energy to be removed = 0.5 * 1100 * 7.16 * 7.16
= 28200 J
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