The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
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Explanation:
Earth rotates in prograde mation.As viewed from the north pole star Polaris.Earth turns counterclockwise,, the north pole is point in the northern,, Hemisphere where Earth's Axis of rotation meets it's surface
Answer:
Based off the word "conserved" I would say
A. Conservation of Momentum.
Explanation:
Answer:
0.9
Explanation:
h = 400 mm, h' = 325 mm
Let the coefficient of restitution be e.
h' = e^2 x h
325 = e^2 x 400
e^2 = 0.8125
e = 0.9
Answer:
3. if you increase your mass you also increase the gravitational pull
6. Radiant energy doesn't depend on a medium and sound energy is dependent on a medium.
Explanation:
i hope this helps-