Given
Car 1
m1 = 1300 kg
v1 = 20 m/s
m2 = 900 kg
v2 = -15 m/s
(Negative sign shows that direction of car 2 is opposite to car 1)
Procedure
As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.
the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period
Thank you
<span>Force = Work done / distance = 4Nm / 2m = 2N</span>
The formula used to find potential energy is <em>P.E. = M * G * H</em> (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is <em>5 * 9.8 * 2</em>, which equals 98.
Answer: The final temperature is 470K
Explanation: Using the relation;
Q= ΔU +W
Given, n = 2mol
Initial temperature T1= 345K
Heat =Q= 2250J
Workdone=W=-870J(work is done on gas)
T2 =Final temperature =?
ΔU =3/2nR(T2-T1)
ΔU=3/2 × 2 ×8.314 (T2 - 345)
ΔU=24.942(T2-345)
Therefore Q = 24.942(T2-345)+ (-870)
2250=24.942(T2-345)+ (-870)
125.09=(T2-345)
T2 =470K
Therfore the final temperature is 470K