Explanation:
When,the vehicle has uniform velocity, it's acceleration becomes zero
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
Kinetic Energy would have to be the answer. Pretty sure.
Answer:
k = 26.25 N/m
Explanation:
given,
mass of the block= 0.450
distance of the block = + 0.240
acceleration = a_x = -14.0 m/s²
velocity = v_x = + 4 m/s
spring force constant (k) = ?
we know,
x = A cos (ωt - ∅).....(1)
v = - ω A cos (ωt - ∅)....(2)
a = ω²A cos (ωt - ∅).........(3)
now from equation (3)
k = 26.25 N/m
hence, spring force constant is equal to k = 26.25 N/m
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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