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3 years ago

WHAT IS THE DENSTY OF ALL THE LAYER IN THE ATMOSPHERE

Physics

2 answers:

Verdich [7]3 years ago

8 0

Answer:

The troposphere starts at the Earth's surface and extends 8 to 14.5 kilometers high (5 to 9 miles). This part of the atmosphere is the most dense. Almost all weather is in this region.

Explanation:

lesantik [10]3 years ago

4 0

<h3><u>Which layer of the atmosphere has the highest density?</u></h3>

The troposphere contains roughly 80% of the mass of Earth's atmosphere. The troposphere is denser than all its overlying atmospheric layers because a larger atmospheric weight sits on top of the troposphere and causes it to be most severely compressed.

The troposphere starts at the Earth's surface and extends 8 to 14.5 kilometers high (5 to 9 miles). This part of the atmosphere is the most dense. Almost all weather is in this region.

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How can you verify the archimedes principle?

Ipatiy [6.2K]

Answer:

It is found that W1 - W2 loss in weight of solid when immersed in water is equal to the weight of the water displaced by the body. This verifies Archimedes' principle.

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3 years ago

Rank the wavelengths of the following quantum particles from the largest to the smallest. If any have equal wavelengths, display

goldfiish [28.3K]

The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.

De Broglie's relationship is given by $\lambda=\frac{h}{mv}$ .

This can be written as $\lambda=\frac{h}{p}$ .....(1) where λ is known as de Broglie wavelength and p is momentum , h = Plank’s constant .

As we know that mass of proton is greater than electron and photon .

(a)For photon , the momentum is given by $p=\frac{E}{c}$ ...(2) where c is the speed is the speed of light .

Putting E = 3eV in equation (2) , we get

$p=\frac{3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^7Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-27}}\\\lambda=4.13\times10^{-7}$

(b) As we know that kinetic energy is given by

$K.E=\frac{1}{2} mv^2\\\\2K.E = mv^2\\2K.E\times m = m^2v^2\\2K.E\times m = (mv)^2\\2K.E\times m = p^2\\\\\sqrt{2K.E\times m }=p$ ...(3)

Where mass of electron is $9.1\times10^-^3^1 kg$ .

Putting K.E = 3eV in equation (3) , we get

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 9.31\times10^{-31} }\\p=\sqrt{89.376\times10^{-50}} \\p=9.45\times10^{-25}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{9.45\times10^{-25}}\\\lambda=0.7005\times10^{-9}$

$\sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 1.67\times10^{-27} }\\p=\sqrt{16.032\times10^{-46}} \\p=4.003\times10^{-23}Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{4.003\times10^{-23}}\\\lambda=1.65\times10^{-11}$

(d) Putting E = 0.3eV in equation (2) , we get

$p=\frac{0.3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^8Js/m$

Putting this value of p in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-28}}\\\lambda=4.13\times10^{-6}$

(e) Putting $p=3eV/c$ in equation (1) , we get

$\lambda=\frac{6.62\times10^{-34}\\\times3\times10^8}{3\times1.6\times10^{-19}}\\\lambda=4.13\times10^{-7}$

On comparing the wavelength order should be (d) > (a) = (e) > (b) > (c) .

Learn more about de brogile here :

brainly.com/question/28165547

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6 0

2 years ago

1. Consider the point exactly halfway between the two wires. Can you adjust the current so that, with current passing through ea

chubhunter [2.5K]

Answer:

hello your question is incomplete attached below is missing part of the question

answer:

1 ) Magnetic field due to long current carrying wire : $B = \frac{U_{0} I}{2\pi d}$

Therefore the net magnetic field due the both wires ; B = B $_{1}$ + B $_{2}$ . when we adjust the current I $_{1}$ = I $_{2}$ then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

$F = \frac{U_{0}I_{1}I_{2} }{2\pi d }$

Explanation:

1 ) Magnetic field due to long current carrying wire : $B = \frac{U_{0} I}{2\pi d}$

Therefore the net magnetic field due the both wires ; B = B $_{1}$ + B $_{2}$ . when we adjust the current I $_{1}$ = I $_{2}$ then the Netfield (B ) = zero

2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires

3) Increase in current through the wire will lead to increase in force and this can be explained via this equation

$F = \frac{U_{0}I_{1}I_{2} }{2\pi d }$

8 0

3 years ago

In 5 or more sentences, explain how you can make everyday count ?

IceJOKER [234]

You can make everyday count by accomplishing all the tasks you have for that day. Also you can try and find the positive throughout the day. Another thing you could do is eat a healthy meal. You can do a workout and that will make you feel better about your self. The last thing you can do is always have a good mindset about that day.

8 0

3 years ago

When a fixed amount of ideal gas goes through an isobaric expansion

Juliette [100K]

Answer:

its temperature must increase.

Explanation:

An isobaric expansion is a transformation in which the pressure of the gas does not change (isobaric), while the volume increases (expansion).

Since the pressure does not change,

"its pressure must increase."

is a false statement.

The 1st law of thermodynamics is

$\Delta U = Q - W$

where

$\Delta U$ is the change in internal energy of the gas, which is proportional to the change in temperature: $\Delta U \propto \Delta T$

Q is the heat supplied to the gas

W is the work done by the gas, which is given by

$W=p\Delta V$

where p is the pressure and $\Delta V$ is the change in volume. Since the gas is expanding, we can say that $\Delta V>0$ , so the gas does positive work:

$W>0$

This means that the option

"the gas does no work."

is false.

Moreover, from the ideal gas law

$pV=nRT$ (2)

we also know that the temperature of the gas is increasing (because p, the pressure, n the number of moles, and R, the gas constant, are all constant in this process, and since the volume V is increasing, than the temperature T must be increasing also)

So, we know that the option

"its internal (thermal) energy does not change. "

is false.

Finally, in an isobaric expansion, in order to keep the pressure constant heat should be supplied to the system, so

"no heat enters or leaves the gas."

is also wrong

We also said from (2) that the temperature of the gas is increasing, therefore the statement

"its temperature must increase."

is the only correct one.

7 0

4 years ago