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ankoles [38]
3 years ago
9

The proper mean lifetime of a muon is 2.2 × 10–6 s. A beam of muons is moving with speed 0.6c relative to an inertial observer.

Approximately how far will a muon in the beam travel, on average, before it decays?

Physics
1 answer:
Viefleur [7K]3 years ago
8 0
This might help you out

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Help mee! Physics i think!
xxTIMURxx [149]

Answer:

Use the form of equation:

Q=mL

You have the specific latent heat of vaporization L = 2.260*10^{6}

And Q, the heat energy supplied, which equals 1695 KJ = 1695*10^{3} J

So you can get the mass by substitution in the formula below.

5 0
3 years ago
Question is down below
rosijanka [135]

The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s

<h3>What are the components of velocity?</h3>

We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.

Hence;

Vertical component = 40 m/s sin 15 degrees = 10.35 m/s

Horizontal component = 40 cos 15 degrees = 38.6 m/s

Learn more about components of velocity:brainly.com/question/14478315

#SPJ1

7 0
2 years ago
Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce
jasenka [17]

Answer:

\lambda= 506.25 nm

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

y=\frac{m \lambda D}{a}

Where:

y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1

Solving for λ:

\lambda=\frac{y*a}{mD}

Replacing the data provided by the problem:

\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm

7 0
3 years ago
Can someone help me find the answer?
Nikitich [7]
Brown black fur and medium tail
8 0
3 years ago
Does mercury have more protons and electrons than tin
Klio2033 [76]
Yes it does.
On the periodic table, tin is #50 and Mercury is # 80. 
3 0
3 years ago
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