Answer:
(a) 2.5 m/s
(b) 37.5 KJ
Explanation:
(a)
From the law of conservation of momentum, Initial momentum=Final momentum

and making
the subject then
and since
is initial velocity of car, value given as 4 m/s,
is the initial velocity of the three cars stuck together, value given as 2 m/s and
is the final velocity which is unknown. By substitution

(b)
Initial kinetic energy is given by

Final kinetic energy is given by

The energy lost is given by subtracting the final kinetic energy from the initial kinetic energy hence
Energy lost=350-312.5=37.5 KJ
Normal force for the rock because that makes an object stable at its position.
static friction because micro-welts hold its particle on its position so it doesn't change in position by a potential energy. Gravity makes it stay on the ground because its force attraction between an object and the earth.
Hope this helps <span />
Answer:
The energy entering, reflecting, absorbed, and emitted by the earth system are the components of the Earth's radiation budget.
Explanation:
I hope this helps also I hope you have a great day and a new year.
Based on my information, this would actually be representing
"the average kinetic energy of water particles". So, as you take notice that where this temperature is being located, and also, how this would be

°C, this would make more sense for this to be representing as <span>the
average kinetic energy of water particles.</span>
Answer:
Explanation:
Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .
velocity of approach = 1.5 - 0 = 1.5
velocity of separation = v₁ + v₂
coefficient of restitution = velocity of separation / velocity of approach
.8 = v₁ + v₂ / 1.5
v₁ + v₂ = 1.2
applying law of conservation of momentum
m x 1.5 + 0 = mv₂ - mv₁
1.5 = v₂ - v₁
adding two equation
2 v ₂= 2.7
v₂ = 1.35 m /s
v₁ = - .15 m / s
During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.
For second collision ,
coefficient of restitution = velocity of separation / velocity of approach
.5 = v₃ + v₄ / 1.35
v₃ + v₄ = .675
applying law of conservation of momentum
m x 1.35 + 0 = mv₄ - mv₃
1.35 = v₄ - v₃
adding two equation
2 v ₄= 2.025
v₄ = 1.0125 m /s
v₃ = - 0 .3375 m / s