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dexar [7]
2 years ago
7

A student pushes a block from rest across a frictionless surface while the block is in front of a motion detector for 3 seconds.

The block first speeds up for the 1st second, then moves at constant velocity for the 2nd second, and finally slows down during the final second.
Which statement best explains the motion of the block using Newton's 1st Law?
There is a net force on the block during the 2nd second as a net force keeps the motion constant.
There is a net force on the block during the 3rd second as a net force keeps the motion constant.
There is no net force on the block during the 2nd second as zero net force keeps the motion constant.
There is no net force on the block during the 3rd second as zero net force causes the motion to accelerate.​
Physics
1 answer:
Furkat [3]2 years ago
5 0

The true statement according to Newton's laws of motion is; "there is no net force on the block during the 2nd second as zero net force keeps the motion constant."

According to the Newton first law of motion, an object will continue in its state of rest or uniform motion unless it is acted upon by a net force. The action of a net force leads to an acceleration.

In the 2nd second of the journey, there is no net force on the block hence the block continues to move at constant velocity.

Learn more: brainly.com/question/13678295

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The work done by a constant force in a rectilinear motion is given by:

W=Fd\cos\theta

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In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

\begin{gathered} W=(17)(12)\cos0+(36)(12)\cos30 \\ W=578.123 \end{gathered}

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Based on the law of conversation of energy how can we reasonably improve a machines ability to do work?
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ivann1987 [24]

<u>Answer:</u>

The final velocity of the two  railroad cars is 1.09 m/s

<u>Explanation:</u>

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by  

\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}

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M1= mass of the first object in kgs = 12000

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V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

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\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}

Which is the final velocity of the two  railroad cars

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