Answer:
The average acceleration of the bearings is 
Explanation:
Given that,
Height = 1.94 m
Bounced height = 1.48 m
Time interval 
Velocity of the ball bearing just before hitting the steel plate
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



Negative as it is directed downwards
After bounce back,
We need to calculate the velocity
Using conservation of energy

Put the value into the formula



We need to calculate the average acceleration of the bearings while they are in contact with the plate
Using formula of acceleration

Put the value into the formula



Hence,The average acceleration of the bearings is 
D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.
Answer:
F = 2 * 30 / 5 = 12 N to stop forward motion
F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees
(12^2 + 16^2)^1.2 = 20 N average force applied
Answer:
K' = 1777.777 J
Explanation:
Given that
m = 40 kg
v= 15 m/s
K=1000
Given that kinetic energy(K) varies with mass(m) and velocity(v)
K= C(mv²)
Where
C= Constant
m=mass
v=velocity
When
m = 40 kg ,v= 15 m/s ,K=1000
K= C(mv²)
1000 = C( 40 x 15²)
C=0.111111
When m = 40 kg and v= 20 m/s
K' = C(mv²)
K= 0.1111 x (40 x 20²)
K' = 1777.777 J
Answer:
t=2.10 s
u= 47.40 m/s
Explanation:
given that
h= 21.8 m
x= 101 m
g=9.8 m/s²
Lets take horizontal speed of ball = u m/s
The vertical speed of the car at initial condition is zero ( v= 0).
We know that

v= 0 m/s

now by putting the values
21.8 = 1/2 x 9.8 x t²
t=2.10 s
This is time when ball was in motion.
Now in horizontal direction
x = u .t
101 = u x 2.1
u= 47.40 m/s