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DENIUS [597]
2 years ago
14

A meter stick A hurtles through space at a speed v = 0.25c relative to you, with its length aligned with the direction of motion

. You stand on Earth and have another meter stick B with its length aligned with the direction of motion of meter stick A. You measure the length of meter stick A to be 1 m. What is the length of meter stick B, L_B, as seen in the rest frame of meter stick A?
Physics
1 answer:
yaroslaw [1]2 years ago
4 0

Answer:

L_0\approx1.0328\ m

Explanation:

Given:

  • relativistic length of stick A, L=1\ m
  • relativistic velocity of stick A with respect to observer, v=0.25c=7.5\times 10^{7}\ m.s^{-1}

<em>Since the object is moving with a velocity comparable to the velocity of light  with respect to the observer therefore the length will appear shorter according to the theory of relativity.</em>

<u> Mathematical expression of the theory of relativity for length contraction:</u>

L=\frac{L_0}{\gamma}

where:

L = relativistic length

L_0= original length at rest

\gamma = Lorentz factor =\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }

\Rightarrow 1=\frac{L_0}{\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }}

L_0=\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }

L_0\approx1.0328\ m

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Answer:

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Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

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Wd = 780 J

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B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

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C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

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Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

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Answer:

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