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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.

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Describe how substance move from one state to another

Mekhanik [1.2K]

The temperature rises until the water reaches the next change of state — boiling. As the particles move faster and faster, they begin to break the attractive forces between each otherand move freely as steam — a gas. The process by which a substance moves from the liquid state to the gaseousstate is called boiling.

4 0

3 years ago

Read 2 more answers

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

2 years ago

A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi

malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$ , where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$

where

$F_{net}$ is the net force

m = 75.0 kg is the mass of the bicyclist

$a=2.20 m/s^2$ is its acceleration

Solving, we find the net force:

$F_{net}=(75.0)(2.20)=165 N$

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

$F_{net}=F-R$

where:

F is the forward force of push

R is the air drag

We know that:

$F_{net}=165 N$ is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

$F=F_{net}+R=165+60=225 N$

6 0

3 years ago

What's the weight of a 30x30x50 cm body with the density of 1.8/cm cube?

grin007 [14]

Answer:

The weight of the body, W = 793.8 m/s²

Explanation:

Given,

The volume of the body, v = 45,000 cm³

The density of the body, ρ = 1.8 g/cm³

The mass of the body is given by the formula,

m = ρ x v

= 1.8 g/cm³ x 45,000 cm³

= 81,000 g

Hence, the mass of the body is m = 81 kg

The weight of the body,

W = m x g

= 81 kg x 9.8 m/s²

= 793.8 m/s²

Hence, the weight of the body, W = 793.8 m/s²

3 0

2 years ago

A block of wood is tied to the bottom of a large container of water so that the block is completely submerged. The density of th

jeka57 [31]

Answer:

58.9 N

Explanation:

The wood is buoyed up by the mass of the water is displaces

wood volume = .030 m^3 = 30 000 cm^3

mass of water displaced = 30 000 g = 30 kg

wood mass = 800 kg/m^3 * .030 m^3 = 24 kg

so tension in string = mg = ( 30 kg - 24 kg) * 9.81 m/s^2 = 58.9 N

6 0

1 year ago

What are muons in cosmology ​

What are muons in cosmology