Answer:
work done is -150 kJ
Explanation:
given data
volume v1 = 2 m³
pressure p1 = 100 kPa
pressure p2 = 200 kPa
internal energy = 10 kJ
heat is transferred = 150 kJ
solution
we know from 1st law of thermodynamic is
Q = du +W ............1
put here value and we get
-140 = 10 + W
W = -150 kJ
as here work done is -ve so we can say work is being done on system
Force acting during collision is internal so momentum is conserve
so (initial momentum = final momentum) in both directions
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.
Let Vx is and Vy are final velocities of car in +x and +y direction respectively.
initial momentum in +ve x (east) direction = final momentum in +ve x direction (east)
- 750*25 + 1150*0 = (750+1150)
Vx
initial momentum in +ve y (north) direction = final momentum in +ve y direction (north)
750*0 - 1150*5 = (750+1150)
Vy
from here you can calculate Vx and Vy
so final velocity V is
<span>V=<span>(√</span><span>V2x</span>+<span>V2y</span>)
</span>
and angle make from +ve x axis is
<span>θ=<span>tan<span>−1</span></span>(<span><span>Vy</span><span>Vx</span></span>)
</span><span>
kinetic energy loss in the collision = final KE - initial KE</span>
Answer:
quantity A is mass and quantity B is wright
Answer:
Option C. 30 m
Explanation:
From the graph given in the question above,
At t = 1 s,
The displacement of the car is 10 m
At t = 4 s
The displacement of the car is 40 m
Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:
Displacement at t = 1 s (d1) = 10 m
Displacement at t= 4 s (d2) = 40
Displacement between t = 1 and t = 4 (ΔD) =?
ΔD = d2 – d1
ΔD = 40 – 10
ΔD = 30 m.
Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.
