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Vlad [161]
3 years ago
13

A muon is a short-lived particle. It is found experimentally that muonsmoving at speed 0.9ccan travel about 1500 meters between

the time they areproduced and the time that they decay.(a) How long does it take the muon to travel this distance?(b) How much time elapses in the muon’s rest frame (the reference frame inwhich the muon is at rest)?
Physics
1 answer:
Monica [59]3 years ago
8 0

Answer:

a)t=5.5\mu s

b)\Delta t'=12.5\mu s

Explanation:

a) Let's use the constant velocity equation:

v=\frac{\Delta x}{\Delta t}

  • v is the speed of the muon. 0.9*c
  • c is the speed of light 3*10⁸ m/s

t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}

t=5.5\mu s

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

\Delta t'=\gamma\Delta t

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

v is the speed of muon (0.9c)

Therefore the time in the rest frame will be:

\Delta t'=\frac{1}{\sqrt{1-\frac{(0.9c)^{2}}{c^{2}}}}\Delta t

\Delta t'=\frac{1}{\sqrt{1-0.9^{2}}}\Delta t

\Delta t'=\frac{1}{0.44}\Delta t

No we use the value of Δt calculated in a)

\Delta t'=2.27*5.5=12.5\mu s

I hope it helps you!

     

 

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The efficiency of the steam engine is 78.9% because the rest of the work input is used to overcome friction.

<h3>What is efficiency of machines?</h3>

Efficiency of a machine expresses the useful work done by a a machine as a percentage.

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The efficiency of machines are always less than 100% percent due to energy losses due to friction and heat.

For the steam engine:

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Efficiency = 225/285 × 100% = 78.9 %

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1 year ago
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mamaluj [8]

Answer:

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7 0
3 years ago
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An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If
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Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

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Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

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350*1.3=850*x

455=850x

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Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?
denpristay [2]

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

\lambda=\frac{2}{n} L


The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m


The wavelength of the 2nd harmonic is:

\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m


The wavelength of the 4th harmonic is:

\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m


It is not possible to find any integer n such that \lambda=5 m, therefore the correct options are A, B and D.

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