Work = Force * distance
Force = 70 N
Work = 3500 J
3500 = 70d
d = 3500/70 = 50 m
Since Pluto is no longer considered a planet, the correct answer would be Neptune. It is eight and farthest known planet of the Solar system. Orbital period of planet Neptune is 168 years. So planet Neptune has completed less than one orbit around the Sun in the last 100 years.
A standing wave is the result of a reflection.
Answer:
particle's potential energy = 70J
Explanation:
From conservation of energy; K1 + Ue1 = K2 + Ue2
where K1 and K2 are the kinetic energies at two positions and Ue1 and Uue2 are the electrical potential energies at two positions.
k1 = 10J, Ue1 = 100J
K2 = 40J
substitute into K1 + Ue1 = K2 + Ue2
Ue2 = K1 + Ue1 - K2
= 10 +100 - 40
Ue2 = 70J
Explanation:
It is given that,
Power of EM waves, P = 1800 W
We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.
Intensity,
![I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BP%7D%7B4%5Cpi%20r%5E2%7D%5C%5C%5C%5CI%3D%5Cdfrac%7B1800%7D%7B4%5Cpi%5Ctimes%20%285%29%5E2%7D%5C%5C%5C%5CI%3D5.72%5C%20W%2Fm%5E2)
The formula that is used to find the rms value of the electric field is as follows :
![I=\epsilon_o cE^2_{rms}](https://tex.z-dn.net/?f=I%3D%5Cepsilon_o%20cE%5E2_%7Brms%7D)
c is speed of light and
is permittivity of free space
So,
![E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m](https://tex.z-dn.net/?f=E_%7Brms%7D%3D%5Csqrt%7B%5Cdfrac%7BI%7D%7B%5Cepsilon_o%20c%7D%7D%5C%5C%5C%5CE_%7Brms%7D%3D%5Csqrt%7B%5Cdfrac%7B5.72%7D%7B8.85%5Ctimes%2010%5E%7B-12%7D%5Ctimes%203%5Ctimes%2010%5E8%7D%7D%5C%5C%5C%5CE_%7Brms%7D%3D46.41%5C%20V%2Fm)
Hence, this is the required solution.