Yes, Sliding friction opposes the movement of the book, slowing it down.sliding That's the 'kinetic' kind.. According to Newton's second law, F=ma. That is, the bear's acceleration should be proportional to the total force acting on the bear. As the bear's velocity is constant, its acceleration is zero. Therefore, the total Force acting on the bear is zero. Thus, the friction has to be equal in magnitude and opposite in direction to the bear's weight. As W=mg, we get that its weight is <span>9.8*400=3,920 Newton. Thus, the friction acting on the bear is 3,920 Newton</span>
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.
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The distance between Mars and the Sun in the scale model would be 1140 m
Explanation:
In this scale model, we have:
represents an actual distance of

The actual distance between Mars and the Sun is 228 million km, therefore

On the scale model, this would corresponds to a distance of
.
Therefore, we can write the following proportion:

And solving for
, we find:

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Answer:
The value of change in internal energy of the gas = + 1850 J
Explanation:
Work done on the gas (W) = - 1850 J
Negative sign is due to work done on the system.
From the first law we know that Q = Δ U + W ------------- (1)
Where Q = Heat transfer to the gas
Δ U = Change in internal energy of the gas
W = work done on the gas
Since it is adiabatic compression of the gas so heat transfer to the gas is zero.
⇒ Q = 0
So from equation (1)
⇒ Δ U = - W ----------------- (2)
⇒ W = - 1850 J (Given)
⇒ Δ U = - (- 1850)
⇒ Δ U = + 1850 J
This is the value of change in internal energy of the gas.