Answer:
I think the answer is B. amount of energy present but I'm not 100% sure
Explanation:
Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height.
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion).
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)?
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m
Work= Force x Distance
Answer: 7500 Joules
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
brainly.com/question/16737526
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