Answer:
Explanation:
Given that :
The radius of the circular loop = 4.0 m
Maximum Emf 
= 5.0 V
The maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies can be determined via the expression;
= 
= 
5.0 = 
5.0 = 
The formula to find work is
. In this case, the force exerted on the rock is 15 Newtons and the distance the rock moved was 2 meters, so 15 * 2 = 30. The answer to your question is
D) 30 n-m.
Answer:
F t = m Δv impulse delivered = change in momentum
Δv = 100 * .1 / .5 = 20 m/s original speed of puck
KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J initial KE of puck
E = μ m g d energy lost by puck
Ff = μ m g = m a deceleration of puck due to friction
a = μ g = 9.8 * .2 = 1.96 m/s^2
v2 = a t + v1 = -1.96 * 4 + 20 = 12.2 m/s speed of puck on striking box
m v2 = M V conservation of momentum when puck strikes box
V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s speed of box after collision
KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J KE of box after collision
KE = μ M g d energy lost by box in sliding distance d
d = 23.3 / (.3 * .8 * 9.8) = 9.91 m distance box slides
