What happens to the gravitation force between two objects that are 15 m apart, when one of them moves 3 m closer?
1 answer:
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Answer:
The value is
Explanation:
From the question we are told that
The diameter of the storm is
The wind speed is
Generally the radius is mathematically represented as
=>
=>
Generally the angular velocity is mathematically represented as
=>
=>
The echo is heard 2.80 s later, this means this is the time the sound takes to travel to the reflecting object and then back to us. So, during this time, the sound wave has covered the distance L between us and the object twice:
The speed of the sound wave is:
, and since it is moving by uniform motion, we can find the distance covered by the wave using
And we said this corresponds to twice the distance between us and the reflecting object, so:
so, the object is 480 meters away.
The speed of the sound wave is:
And we said this corresponds to twice the distance between us and the reflecting object, so:
so, the object is 480 meters away.
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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