Answer:
J = 0.7 N-s away from wall
Explanation:
Given that,
Mass of the ball, m = 0.2 kg
Initial speed of the ball, u = 2 m/s
Final speed of the ball, v = -1.5 m/s (as it rebounds)
Impulse is equal to the change in momentum. So,

So, the impulse given to the ball is 0.7 N-m and it is away from the wall. Hence, the correct option is (c).
Too easy way too easy I mean it’s easy
Answer:
Static force of friction = 428.75 N
Kinetic Force of Friction = 281.75 N
Explanation:
The static force of friction is equal to Coefficient of static Friction * Mass * g
= 0.35 *125 * 9.8
= 428.75 N
The Kinetic force of friction is equal to Coefficient of kinetic Friction * Mass * g
= 0.23 *125 * 9.8
= 281.75 N
Answer:


Explanation:
Given that.
Force acting on the particle, 
Position of the particle, 
To find,
(a) Torque on the particle about the origin.
(b) The angle between the directions of r and F
Solution,
(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :




So, the torque on the particle about the origin is (32 N-m).
(b) Magnitude of r, 
Magnitude of F, 
Using dot product formula,




Therefore, this is the required solution.
Answer:
76969.29 W
Explanation:
Applying,
P = F×v............. Equation 1
Where P = Power, F = force, v = velocity
But,
F = ma.......... Equation 2
Where m = mass, a = acceleration
Also,
a = (v-u)/t......... Equation 3
Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s
Substitute these values into equation 3
a = (12.87-0)/3.47
a = 3.71 m/s²
Also Given: m = 1612 kg
Substitute into equation 2
F = 1612(3.71)
F = 5980.52 N.
Finally,
Substitute into equation 1
P = 5980.52×12.87
P = 76969.29 W