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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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6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final
Alenkasestr [34]

Answer:

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy =  ¹/₂m₁u₁² + ¹/₂m₂u₂²  = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

                                                               = ¹/₂ x (9.43)²(25+6.8)

                                                               = 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J

                               

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3 years ago
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Answer: The answer is D.

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Covert 60 mph to SI mks units
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26.82m/s

Explanation:

Given:

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To solve this problem, we have to find the right and appropriate conversion factor which equals to 1 to multiply this unit with:

   we are converting:

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Now to convert from mph to m/s

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learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

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