Answer:
the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s
Explanation:
given information
car's mass, m = 1200 kg
= 100 m
= 
= 150 m
= 0
according to conservative energy
the distance from point A to B, h = 150 m - 100 m = 50 m
the initial speed 
final speed
= 0
thus,
² =
² - 2 g h
0 =
² - 2 g h
² = 2 g h
= √2 g h
= √2 (9.8) (50)
= 31.3 m/s
Answer: :) i think its the t
Answer:
a) The potential energy in the system is greatest at X.
Explanation:
Let be X the point where a ball rests at the top of a hill. By applying the Principle of Energy Conservation, the total energy in the physical system remains constant and gravitational potential energy at the top of the hill is equal to the sum of kinetic energy, a lower gravitational energy and dissipated work due to nonconservative forces (friction, dragging).

Conclusions are showed as follows:
a) The potential energy in the system is greatest at X.
b) The kinetic energy is the lowest at X and Z.
c) Total energy remains constant as the ball moves from X to Y.
Hence, the correct answer is A.
Answer:
Explanation:
we know that the time rate of change of velocity is called acceleration mathmetically;
a=Δv/Δt
Δv=a*Δt
a=10 m/s²
t=10 sec
Δv=10*10=100 m/s
Answer:
The change in the mechanical energy of the projectile is 43,750 J
Explanation:
Given;
mass of the projectile, m = 5 kg
initial velocity of the projectile, u = 200 m/s
final velocity of the projectile, v = 150 m/s
The change in mechanical energy is calculated from the principle of conservation of energy;
ΔP.E = ΔK.E
The change in potential energy is zero (0)
0 = ΔK.E
ΔK.E = K.E₁ - K.E₂
ΔK.E = ¹/₂mu² - ¹/₂mv²
ΔK.E = ¹/₂m(u² - v²)
ΔK.E = ¹/₂ x 5(200² - 150²)
ΔK.E = 43,750 J
Therefore, the change in the mechanical energy of the projectile is 43,750 J