#12 and 15 PHYS 120 11. What is the magnitude of the velocity of 14. If the elev
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#12 and 15
PHYS 120 11. What is the magnitude of the velocity of 14. If the elevator is moving upward with a constant velocity of u = 80, what is the the man after he pulls on the sled? reading on the scale? (a) o.0m (b) 4.0 (c) 8.0 (d) None of the above (a) 0.0N (b) 740N (c) 1400/N (d) None of the above For the following two questions con- sider: A woman pulling a cart which has frictionless bearings between the axels and the wheels while she is walking. 15. Now the elevator slows to a stop with an acceleration of magnitude a = 0.35g What is the reading on the scale while the elevator is slowing down? (a) 0.0N, because the man is no longer in 12. What is the relative force experianced by the woman? (a) She feels more force than that which contact with the scale. is being applied to the cart. (b) She feels less force than that which is ( (b) 990N 740N being applied to the cart. (c) She feels the same amount of force as (d) 480N that which is being applied to the cart. For the following two questions con- sider: Two blocks connected by a rope. Block one has massm 2.5kg, and block two has mass m2 =/4.0kg./ Block (d) None of the above. 13. If the woman applies a force of 150N to the two is being pulled by a rope which is at- rope, and she has a mass of 50.kg, what is the coefficient of static friction between her tached on the opposite side of block two as block one. Let the tension in the rope be T 25N, the coefficient of kinetic friction be '= 0.25, and use g 10% shoes and the ground? (a) 0.31 (b) 3.0 (c) 0.50 (d) There is no force of static friction. cS 16. What is the acceleration of mass one? (b) 3.8 (c) 16 (d) None of the above. Only a force of kinetic friction. For the following two questions con- sider: A man with mass m 75kg is standing on a scale in an elevator.Explanation / Answer
(12).Option (B) is correct
Since she feels only a force of magnitude along motion direction
(15).mass m=75kg
Acceleration a=0.35g
Reading on the scale R=m(g-a)
R=75(0.65g)
=75(0.65)(9.8)
=477.75N=480N
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