Answer:
Explanation:
Given that
x= 150 ft
y= 14 ft
From the diagram
When ,x= 150 ft and y= 14 ft
z=150.74 ft
By differentiating with respect to time t
Here x is constant that is why
Now by putting the values in the above equation we get
Therefore the distance between balloon and observer increasing with 0.65 ft/s.
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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Answer:
How does newtons first two laws of motion apply to the toy car?
Explanation:
Answer:
I hope it's helpful for you ☺️
Explanation:
i hope this helps, its not the same person but its the same equation.