It is wasted, most likely as light, in this case, or it is lost during the transport of electricity.
Initial velocity u = 40
Angle at launch = 55 degrees
At maximum height v = 0, velocity equation v^2 = u^2 - 2gh,
0 = (40 x sin55)^2 - 2 x 9.81 x h => (40 x 0.819)^2 = 19.62h => h = 32.76^2 /
19.62
Maximum height = 54.7 m
We have v = u - gt => 0 = (40 x sin55) - 9.81 x t => t = 32.76 / 9.81 => t =
3.34 s
Time taken to hit the ground is 2t = 2 x 3.34 = 6.68 s
Distance from castle to trebuchet = utcos55 = 40 x 6.68 x 0.573 = 153.1 m
Answer:
Work done is 12.3 J
Explanation:
We have,
Mass of puck, m = 0.35 kg
Force of friction acting on the puck when it slides is 0.15 N
Distance travelled by the puck is 82 m.
It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by

The work done on the puck is 12.3 J.
Answer:
The dimension is 
Explanation:
From the question we are told that

Here ![[J] = \frac{1}{L^2 T}](https://tex.z-dn.net/?f=%5BJ%5D%20%3D%20%5Cfrac%7B1%7D%7BL%5E2%20T%7D)
![[n] =\frac{1}{L^3}](https://tex.z-dn.net/?f=%5Bn%5D%20%3D%5Cfrac%7B1%7D%7BL%5E3%7D)
![[x] = L](https://tex.z-dn.net/?f=%5Bx%5D%20%3D%20L)
So
![\frac{1}{L^2 T} = -D \frac{d(\frac{1}{L^3})}{d[L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BL%5E2%20T%7D%20%3D%20%20-D%20%5Cfrac%7Bd%28%5Cfrac%7B1%7D%7BL%5E3%7D%29%7D%7Bd%5BL%5D%7D)
Given that the dimension represent the unites of n and x then the differential will not effect on them
So
=> 
=> 
Answer:
700ma
Explanation:
all u do is have to add them up.