Answer:
Power = 0.33 Watts
Explanation:
Given the following data;
Distance = 1m
Force = 20N
First of all, we would solve for the work done by the boy.
Workdone = force * distance
Substituting into the equation, we have;
Workdone = 20*1 = 20J
Now to find power;
Power = workdone/time
Power = 20/60
Power = 0.33 Watts.
Explanation:
The electric field at a distance r from the charged particle is given by :
![E=\dfrac{kq}{r^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bkq%7D%7Br%5E2%7D)
k is electrostatic constant
if r = 2 m, electric field is given by :
![E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)](https://tex.z-dn.net/?f=E_1%3D%5Cdfrac%7Bkq%7D%7B%282%29%5E2%7D%5C%5C%5C%5C%3D%5Cdfrac%7Bkq%7D%7B4%7D%5C%20.....%281%29)
If r = 1 m, electric field is given by :
![E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)](https://tex.z-dn.net/?f=E_2%3D%5Cdfrac%7Bkq%7D%7Br_2%5E2%7D%5C%5C%5C%5C%3D%5Cdfrac%7Bkq%7D%7B1%7D%5C%20....%282%29)
Dividing equation (1) and (2) we get :
![\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1](https://tex.z-dn.net/?f=%5Cdfrac%7BE_1%7D%7BE_2%7D%3D%5Cdfrac%7B%5Cdfrac%7Bkq%7D%7B4%7D%7D%7Bkq%7D%5C%5C%5C%5C%5Cdfrac%7BE_1%7D%7BE_2%7D%3D%5Cdfrac%7B1%7D%7B4%7D%5C%5C%5C%5CE_2%3D4%5Ctimes%20E_1)
So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.
Answer:
1) It expresses the rate (top speed) at which it can move with time.
2) P = 20 W
3) h = 18 km
Explanation:
1) Power is the rate of transfer of energy.
⇒ Power = ![\frac{Energy(or workdone)}{Time}](https://tex.z-dn.net/?f=%5Cfrac%7BEnergy%28or%20workdone%29%7D%7BTime%7D)
i.e P = ![\frac{E}{t}](https://tex.z-dn.net/?f=%5Cfrac%7BE%7D%7Bt%7D)
Thus a car's engine power is 44000W implies that the engine of the car can propel the car at this rate. This expresses the rate (top speed) at which it can move with time.
2) m = 400g = 0.4 kg
t = 20 s
h = 100m
g = 10 m/![s^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D)
P = ![\frac{mgh}{t}](https://tex.z-dn.net/?f=%5Cfrac%7Bmgh%7D%7Bt%7D)
= ![\frac{0.4*10*100}{20}](https://tex.z-dn.net/?f=%5Cfrac%7B0.4%2A10%2A100%7D%7B20%7D)
= ![\frac{400}{20}](https://tex.z-dn.net/?f=%5Cfrac%7B400%7D%7B20%7D)
P = 20 W
3) u = 600 m/s
g = 10 m/![s^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D)
From the third equation of free fall,
=
- 2gh
V is the final velocity, U is the initial velocity, h is the height.
0 =
- 2 x 10 x h
0 = 360000 - 20h
20h = 360000
h = ![\frac{360000}{20}](https://tex.z-dn.net/?f=%5Cfrac%7B360000%7D%7B20%7D)
= 18000
h = 18 km
The maximum height of the bullet would be 18 km.