Answer:
The football leaves with the velocity, u = 15.68 m/s
Explanation:
Given data,
The football bounces back up off the ground and is airborne for, t = 3.2 s
Let the football bounces back up off the ground in the vertical direction
The formula for time of flight is given by,
t = 2u /g
∴ u = gt / 2
Substituting the values,
u = 9.8 x 3.2 / 2
u = 15.68 m/s
Hence, the football leaves with the velocity, u = 15.68 m/s
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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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We can use renewable sources
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There are 5,280 feet in 1 mile.
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