Which is equal to 4 hectometers?

The is greater for low-mass stars than it is for high-mass stars. 2. The stars known as are the very largest and brightest of al

Anvisha [2.4K]

Answer:

red giant stars/ red hyper giants

Explanation:

take stephenson 2-18 for example the star is only 3200k where the sun is around 5000k.

3 0

3 years ago

A hollow, uniformly charged sphere has an inner radius of r1 = 0.105 m and an outer radius of r2 = 0.31 m. The sphere has a net

Sliva [168]

Answer:

E = 77532.42N/C

Explanation:

In order to find the magnitude of the electric field for a point that is in between the inner radius and outer radius, you take into account the Gauss' law for the electric flux trough a spherical surface with radius r:

$\int E\cdot dS=\frac{Q}{\epsilon_o}$ (1)

Q: net charge of the hollow sphere = 1.9*10-6C

ε0: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

Furthermore, you have that the net charge contained in a sphere of radius r is:

$Q=\rho V=\rho \frac{4\pi (r^3-r_1^3)}{3}$ (2)

with the charge density is:

$\rho=\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}$ (3)

r2: outer radius = 0.31m

r1: inner radius = 0.105m

The electric field trough the Gaussian surface is parallel to the normal to the surface, the, you have in the integral of the equation (1):

$\int E\cdot dS=E(4\pi r^2)$ (4)

where you have used the expression for a surface of a sphere.

Next, you replace the expressions of equations (2), (3) and (4) in the equation (1) and solve for E:

$E(4\pi r^2)=\frac{1}{\epsilon_o}\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}(\frac{4\pi (r^3-r_1^3)}{3})\\\\E=\frac{1}{\epsilon_o}\frac{Q(r^3-r_1^3)}{4\pi r^2(r_2^3-r_1^3)}$