Answer:
static coefficient = 0,203 & kinetic coefficient = 0,14
Explanation:
There are two (2) conditions, when the desk is about to move and when the desk is moving. In the attachements you can see the two free body diagram for each condition.
In the first condition, there is no movement and the force is 12 N, in the image we can see the total forces are equal to 0 and by the definition of the friction force we can get the static friction coefficient.
In the second condition there is movement in the direction of the force which is equal to 8 N, again by the definition of the friction force we can get the kinetic friction coefficient. Since the desk is moving with constant velocity there is not acceleration.
Answer:
(A.)Nuclear fission and beta decay (electron emission)
Answer:
306 m/s
Explanation:
Law of conservation of momentum
m1v1 + m2v2 = (m1+m2)vf
m1 is the bullet's mass so it is 0.1 kg
v1 is what we're trying to solve
m2 is the target's mass so it is 5.0 kg
v2 is the targets velocity, and since it was stationary, its velocity is zero
vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s
plugging in, we get
(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)
(0.1)(v1) + 0 = 30.6
(0.1)(v1) = 30.6
v1 = 306 m/s