Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by

where k is the elastic constant of the spring and

is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:

where the negative sign is given by the fact that

points in the opposite direction of the displacement of the cart, and where

therefore, the work done by the weight is
Answer:
Explanation:
mass of the ball = 146 g = 146 / 1000 = 0.146 kg
initial speed of the ball = 40.6 m/s
final speed of the ball = - 45.1 m/s
time of impact = 1.05 ms = 1.05 / 1000 = 0.00105 s
impulse, Ft = change in momentum = mv - mu = m (v-u)
F = m (v - u) / t = 0.146 kg ( -45.1 -40.6) / 0.00105 s = -11916.4 N
Answer:
B, B (decreases, a clockwise)
Explanation:
Finally, the switch on the electromagnet is reopened. The magnitude of the external magnetic flux through the wire loop <u>decreases</u>, and there is <u>a clockwise</u>, current induced in the loop (as seen from the left).