Answer:
(a) 8.117 (b) 0.742
Explanation:
We have given distance s =40 m time t=7.5 sec
Final velocity v =2.55 m/sec
From the first equation of motion 
(negative sign because there is retrdation as the truck speed is slowing down )
So 
--------------eqn 1
From the second equation of motion 
( negative sign because there is retrdation as the truck speed is slowing down )
So 
------------------eqn 2
On solving eqn1 and eqn 2
u=8.117 m/sec and a=-0.742
Answer:
It depends how far the monkey is and how hard you throw the object because If the monkey is close, then you will most likely hit it
Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate doesn't move (static), acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = Fm - Ff = 0.
Fm is the applied force
Ff is the frictional force
Since Fm - Ff = 0
Fm = Ff
This means that the applied force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
R = 31.2 × 9.8
R = 305.76N
From the formula
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
Answer:
Option A is the correct answer.
Explanation:
Here momentum is conserved.
That is 
Substituting values
Speed of block A after collision = 10 m/s
Option A is the correct answer.
