What is an intraplate quake? What causes them?
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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).
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The vertical component of force exerted by the hi.nge on the beam will be,142.10N.
To find the answer, we need to know more about the tension.
<h3>How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>- Let's draw the free body diagram of the system.
- To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.
- To find the answer, we have to find the tension,
- Thus, the vertical component of the force exerted by the hi.nge on the beam will be,
Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.
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