In order:
Theory
Fact
Hypothesis
The answer to this question is false
<h2>
Its velocity when it crosses the finish line is 117.65 m/s</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = ?
Time, t = 6.8 s
Displacement, s = 1/4 mi = 400 meters
Substituting
s = ut + 0.5 at²
400 = 0 x 6.8 + 0.5 x a x 6.8²
a = 17.30 m/s²
Now we have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = ?
Time, t = 6.8 s
Acceleration, a = 17.30 m/s²
Substituting
v = u + at
v = 0 + 17.30 x 6.8
v = 117.65 m/s
Its velocity when it crosses the finish line is 117.65 m/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
#SPJ9
