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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When two different air masses meet, a boundary is formed. the boundary between two air masses is called a front. weather at a front is usually cloudy and stormy. there at four different fronts: cold, warm, stationary, and occluded
Answer:
the small one will have greater acceleration
Explanation:
force and mass are inversely proportional. force and acceleration are directly proportional. which means greater mass have smaller acceleration and smaller mass has greater acceleration. this is due to newtons second law of motion.
The object is moving with constant speed means there is no change in speed Hence the acceleration is 0
Where, u is initial speed,
v is final speed
t is time taken and
a is acceleration
a = ( v-u) /t
a =(0)/t because u=v
a=0
Therefore, Acceleration is 0