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).
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Answer:
6858.5712 m/s
Explanation:
Given that:
Radius, r
R = 3.20 * 10^3.
Normal force = 0.5 * normal weight
Normal force = Fn ; Normal weight = Fg
Fn = 0.5Fg
Recall:
mv² / R = Fn + Fg
Fn = 0.5Fg
mv² / R = 0.5Fg + Fg
mv² /R = 1.5Fg
mv² = 1.5Fg * R
F = mg
mv² = 1.5* mg * R
v² = 1.5gR
v = sqrt(1.5gR)
V = sqrt(1.5 * 9.8 * 3.2 * 10^3)
V = sqrt(47.04^3)
V = 6858.5712 m/s
The first one, as the mass is higher so it accelerates more
Answer:
108 km
Explanation:
The conversion factor between meters and feet is
1 m = 3.28 ft
So the second altitude, written in feet, can be rewritten in meters as

or in kilometers,

the first altitude in kilometers is

so the difference between the two altitudes is
