Answer:

Explanation:
The proton is under a linear motion with constant acceleration. So, we use the kinemtic equations to calculate its final speed. We know its acceleration, its initial speed and its traveled distance. Thus, we use the following equation:

A real question, one with a question mark?
Answer:
(E)56.0 m/s
Explanation:
Height =h=-160 m
Because the wallet moving in downward direction
Time=t=7 s
Final speed of wallet=v=0
We have to find the speed of helicopter ascending at the moment when the passenger let go of the wallet.

Where 
Substitute the values



Option (E) is true
Answer:
v=0.94 m/s
Explanation:
Given that
M= 5.67 kg
k= 150 N/m
m=1 kg
μ = 0.45
The maximum acceleration of upper block can be μ g.
a= μ g ( g = 10 m/s²)
The maximum acceleration of system will ω²X.
ω = natural frequency
X=maximum displacement
For top stop slipping
μ g =ω²X
We know for spring mass system natural frequency given as

By putting the values

ω = 4.47 rad/s
μ g =ω²X
By putting the values
0.45 x 10 = 4.47² X
X = 0.2 m
From energy conservation


150 x 0.2²=6.67 v²
v=0.94 m/s
This is the maximum speed of system.
It should be C. If the object is denser than the fluid, it will sink. If it isn't, it will float