(a)
KE = m v^2 / 2 = (1200 kg)(20 m/s)^2 / 2 = 240,000 J
(b)
The energy is entirely dissipated by the force of friction in the brake system.
(c)
W = delta KE = KEf - KEi = (0 - 240,000) J = -240,000 J
(d)
Fd = delta KE
F = (delta KE) / d = (-240,000 J) / (50 m) = -4800 N
The magnitude of the friction force is 4800 N.
Answer:
Approximate height of the building is 23213 meters.
Explanation:
Let the height of the building be represented by h.
0.02 radians = 0.02 × 
= 0.02 x (180/
)
0.02 radians = 1.146°
10.5 km = 10500 m
Applying the trigonometric function, we have;
Tan θ = 
So that,
Tan 1.146° = 
⇒ h = Tan 1.146° x 10500
= 2.21074 x 10500
= 23212.77
h = 23213 m
The approximate height of the building is 23213 m.
I believe the answer is H for when you bounce it, it has stress when it hits the floor and then goes up giving it kinetic