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MA_775_DIABLO [31]
3 years ago
12

A rod with a rest length of 1m whizzes past an observer at a speed of 0.995c, where c is the speed of light. Is it possible that

the observer could also measure the length of the moving rod to be 1 m? A. yes B. no
Physics
2 answers:
Tresset [83]3 years ago
6 0

Answer:

B. no

Explanation:

  • When any body moves at a speed comparable to the speed of light (<em>i.e. relativistic speed</em>) then the observer sees a contraction in length of the body along the axis of motion.

Assuming that the motion of the body is along the axis of the rod, then the observer will measure its length to be lesser than its length at rest.

<u>Then according to Einstein's theory of relativity:</u>

L=\frac{L_0}{\gamma}

where:

L_0=  original length of the object (along the direction of motion)

L = observed length of the rod

\gamma = Lorentz factor

\gamma\equiv \frac{1}{\sqrt{1-\frac{v^2}{c^2} } }

abruzzese [7]3 years ago
6 0

Answer:

B. no

Explanation:

If any body moves at a rate equal to the speed of light (i.e. relativistic velocity), the observer experiences a contraction along the direction of motion in the span of the body.

Lets us suppose that the motion of the body is along the axis of the rod, the observer will measure the length of the rod which will be lesser than its length in rest position.

And contracted length is given by

L=L_o\sqrt{1-\frac{v^2}{c^2} }

Lo= original length of the object along its axis.

L= length measured by the observer.

v= 0.995c

c= speed of light

So,observer will measure a length less than 1 m

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oee [108]

Answer:

A. Their shape

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A(n) ________ microscope keeps an object in focus when the objective lens is changed.
Anvisha [2.4K]

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5)A 0.50 kg hockey puck is at rest on ice when you hit it with a hockey stick, applying a force of 100 N for
mojhsa [17]

Answer:

F t = m Δv         impulse delivered = change in momentum

Δv = 100 * .1 / .5 = 20 m/s     original speed of puck

KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J     initial KE of puck

E = μ m g d        energy lost by puck

Ff = μ m g = m a      deceleration of puck due to friction

a = μ  g = 9.8 * .2 = 1.96 m/s^2

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m v2 = M V       conservation of momentum when puck strikes box

V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s     speed of box after collision

KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J     KE of box after collision

KE = μ M g d     energy lost by box in sliding distance d

d = 23.3 / (.3 * .8 * 9.8) = 9.91 m     distance box slides

7 0
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dusya [7]

Answer:

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