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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no it doesn't why because I think that it is not the same but different.
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19.8 N force is tending to lift Rover vertically off the ground.
<h3>What is horizontal and vertical component?</h3>
The horizontal velocity component (
) describes the influence of the velocity in displacing the projectile horizontally. The vertical velocity component (
) describes the influence of the velocity in displacing the projectile vertically.
According to the question,
The women pulls the dog with a force of 30 N at an angle of 29° from the horizontal.
Horizontal component= 30cos(29°) = 22.2 N
Vertical component = 30sin(29°) = 19.8 N
Therefore,
The horizontal component would tend to make the dog move forward and the vertical component would tend lift it off the ground.
Hence,
19.8 N force is tending to lift Rover vertically off the ground.
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Answer:
Hope it helps..
Explanation:
Let n be the number of the vernier scale division which coincides with the main scale division. Rotate the vernier caliper 90° and repeat the steps 4 and 5 for measuring the internal diameter in perpendicular direction. To measure the depth, find the total reading and zero correction.
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