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pashok25 [27]
3 years ago
8

An object of mass 100 kg is initially at rest on a horizontal frictionless surface. At time t = 0, a horizontal force of 10 N is

applied to the object for 1 s and then removed. Which of the following is true of the object at time t = 2s if it is still on the surface?
(A) It is at the same position it had at t 0, since a force of 10 N is not large enough to move such a massive object.
(B) It is moving with constant nonzero acceleration.
(C) It is moving with decreasing acceleration.
(D) It is moving at a constant speed
(E) It has come to rest some distance away from the position it had att 0.
Physics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

(D) It is moving at a constant speed

Explanation:

Before t = 1s. Due to the force, albeit small, acting on the object, since there's no static friction stopping the object from moving, this mass object would have a constant acceleration and it's velocity would be increasing.

According to Newton's 1st law, an object will stay at a constant speed if the net force acting on it is 0. After t = 1s, horizontally speaking there's no other force exerting on the mass object. There is no friction force at play here as the surface is frictionless.

Therefore the correct statement is (D) It is moving at a constant speed

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True.

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Which one of the following is a derived SI unit?<br>A.newton B.meter C.mole d.Kilogram ​
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3 years ago
Hannah walks 0.30 km to class in 5.0 min. what is her average speed in m/s?
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An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno
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Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

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Explanation:

   From the question we are told that

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              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

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           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

             \epsilon =0.11232 V

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