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diamong [38]
3 years ago
10

Consider the four masses. Describe the motion of the masses if you used THE SAME amount of force to move each one over THE SAME

time.
A) The greatest mass would not move at all.
B) All four masses would move the same distance.
C) The lower the mass, the greater the distance the mass would move.
D) The greater the mass, the greater the distance the mass would move.

Physics
2 answers:
ahrayia [7]3 years ago
7 0

Answer: Option (C) is the correct answer.

Explanation:

When same amount of force is applied to all the given masses then object with smaller mass will move to the greatest distance.

Because smaller is the mass small will be the force required to displace the object.

Thus, we ca conclude that out of the given options, the lower the mass, the greater the distance the mass would move.

Illusion [34]3 years ago
4 0
The answer is c. if itis heavier, u have to push hardier or it to move the same distance. make sense??
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Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

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  • direction of velocity of jet, east relative to the ground
  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

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Answer:

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Explanation:

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X axis

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