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saul85 [17]
3 years ago
13

A rocket passes you at a speed of 0.85c, and you measure its length to be 35.0 m. What is its measured length when at rest?

Physics
1 answer:
viktelen [127]3 years ago
8 0

Answer:

66.4 m

Explanation:

To solve the problem, we can use the length contraction formula, which states that the length observed in the reference frame moving with the object (the rocket) is given by

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

where

L_0 is the proper length (the length measured from an observer at rest)

v is the speed of the object (the rocket)

c is the speed of light

Here we know

v = 0.85c

L = 35.0 m

So we can re-arrange the equation to find the length of the rocket at rest:

L_0 = \frac{L}{\sqrt{1-(\frac{v}{c})^2}}=\frac{35.0}{\sqrt{1-(\frac{0.85c}{c})^2}}=66.4 m

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A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when
Marianna [84]

Answer:

A) P=13.92\ J.s^{-1}

B) v=3730.9912\ m.s^{-1}

C) v=74.44\ mm.s^{-1}

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

  • Distance form the sound source, s=50\ m
  • sound intensity level at the given location, \beta=114\ dB
  • diameter of the eardrum membrane in humans, d=8.4 \times 10^{-3}\ m
  • We have the minimum detectable intensity to the human ears, I_0=10^{-12}\ W.m^{-2}

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

\beta=10\ log(\frac{I}{I_0} ) ..........................................(1)

114=10\ log\ (\frac{I}{10^{-12}} )

11.4=log\ I+12\ log\ 10

I=0.2512\ W.m^{-2}

<em>As we know :</em>

I=\frac{P}{A}

0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }

P=13.92\ J.s^{-1} is the energy transferred to the  eardrums per second.

(B)

mass of mosquito, m=2\times 10^{-6}\ kg

<u>Now the velocity of mosquito for the same kinetic energy:</u>

KE=\frac{1}{2} m.v^2

13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2

v=3730.9912\ m.s^{-1}

(C)

Given:

  • Sound intensity, \beta = 20\ dB

<u>Using eq. (1)</u>

20=10\ log\ (\frac{I}{10^{-12}} )

2=log\ I+12\ log\ 10

I=10^{-10}\ W.m^{-2}

Now, power:

P=I.A

P=10^{-10}\times \pi\times \frac{8.4^2}{4}

P=5.54\times 10^{-9}\ J.s^{-1}

Hence:

KE=\frac{1}{2} m.v^2

5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2

v=0.07444\ m.s^{-1}

v=74.44\ mm.s^{-1}

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0
3 years ago
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