Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :
Work done on the wagon is given by the product of net force and displacement. It is given by :
W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.
Answer:
300 m
Explanation:
The train accelerate from the rest so u = 0 m/sec
Final speed that is v = 80 m/sec
Time t = 30 sec
The distance traveled by first plane = 1200 m
We know the equation of motion 
where s is distance a is acceleration and u is initial velocity
Using this equation for first plane 
As the acceleration is same for both the plane so a for second plane will be 2.67 
The another equation of motion is 
using this equation for second plane 
s = 300 m
1. The velocity decreases, and the kinetic energy decreases.
2. An increase in temperature difference between the inside and outside of the building.
3. The total kinetic energy remains the same.
4. 76,761 J
5. The energy loss must increase.
The speed of light generally would be 300000km/s but since the train is moving in the same direction as the light it would apparently appear to be 100000km/s
S = ut + 0.5at^2
<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} </span>
<span>t^2 ~ 2.04 </span>
<span>t ~ 1.43 seconds</span>
