All categories

3 years ago

What are muons in cosmology

Physics

1 answer:

yanalaym [24]3 years ago

4 0

Answer:

Explanation:

Muons in the atmosphere, a component of cosmic rays. Atmospheric muons are an essential component of cosmic ray showers. When a high energy primary particle coming from space collides with a nucleus of the upper atmosphere, it generates a spray of particles which later interact in their turn.

You might be interested in

Two particles with the same charge will repel each other. True or false

FinnZ [79.3K]

True. Think of a magnet and how they only connect to the opposite charges.

7 0

3 years ago

How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water

alekssr [168]

Answer

A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.

5 0

3 years ago

The correct answer is 40 or - 40??

arsen [322]

Answer:

Explanation:

6 0

3 years ago

List down different layers of the sun. Rank these layers based on their distance from the sun’s center

vfiekz [6]

Answer:

Core

Radiative zone

Convective zone

Photosphere

Chromosphere

Transient region

Corona

Ranks of layers based on their distance from the sun’s center

1st-corona

2nd-Transient region

3rd-chromosphere

4th-Photosphere

5th-convective zone

6th-radiative zone

7th-core

6 0

3 years ago

Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$