If its uniform there is no force.
C) 0 N
Answer:
you can't
Explanation:
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Three points affecting are :
a. Magnitude of force.
b. Distance of point of contact.
c. Angle at which force is applied.
Answer:
(a) σ = 3.41*10⁻7C/m^2
(b) E = 38,530.1 N/C
Explanation:
(a) In order to calculate the resulting surface charge density, you use the following formula:
(1)
σ: surface charge density
Q: charge of the satellite = 3.1 µC = 3.1*10^-6C
S: surface area of the satellite
The satellite has a spherical form, then, the area of the surface is given by:
(2)
r: radius of the satellite = d/2 = 1.7m/2 = 0.85m
You replace the equation (2) into the equation (1) and solve for the surface charge density:
![\sigma=\frac{3.1*10^{-6}C}{4\pi (0.85m)^2}=3.41*10^{-7}\frac{C}{m^2}](https://tex.z-dn.net/?f=%5Csigma%3D%5Cfrac%7B3.1%2A10%5E%7B-6%7DC%7D%7B4%5Cpi%20%280.85m%29%5E2%7D%3D3.41%2A10%5E%7B-7%7D%5Cfrac%7BC%7D%7Bm%5E2%7D)
The surface charge density acquired by the satellite on one orbit is 3.41*10⁻7C/m^2
(b) The electric field just outside the surface is calculate d by using the following formula:
(3)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
R: radius of the satellite = 0.85m
![E=(8.98*10^9Nm^2/C^2)\frac{3.1*10^{-6}C}{(0.85m)^2}=38530.1\frac{N}{C}](https://tex.z-dn.net/?f=E%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5Cfrac%7B3.1%2A10%5E%7B-6%7DC%7D%7B%280.85m%29%5E2%7D%3D38530.1%5Cfrac%7BN%7D%7BC%7D)
The magnitude of the electric field just outside the sphere is 38,530.1 N/C
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