A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc
2 answers:
Answer:
Choice a. , assuming that the skating rink is level.
Explanation:
<h3>Net force in the horizontal direction</h3>There are two horizontal forces acting on the boy:
- The pull of his friend, and
- Frictions.
The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.
The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.
.
The net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.
However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.
<h3>Net force</h3>Therefore, the (combined) net force on this boy would be:
.
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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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