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2 years ago

Please help me with my question

Physics

1 answer:

Mashutka [201]2 years ago

3 0

Answer:

Brownian motion and diffusion

Explanation:

Brownian motion: taking into consideration, SMOKE. when a burning fire is kindled, smoke is seen to move in the way or direction that is not straight but keep changing continuously ( in haphazard manner). Haphazard manner of smoke is an evidence that matter are made of molecules and are constantly in motion

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Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many

Zielflug [23.3K]

Answer:

The $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

Explanation:

Given :

Current $I = 1 \times 10^{5}$ A

Charge of electron $e = 1.6 \times 10^{-19}$ C

Time $t = 1$ sec

From the formula of current,

Current is the number of charges flowing per unit time.

$I = \frac{ne}{t}$

Where $n =$ number of charges means in our case number of electrons

$n = \frac{It}{e}$

$n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }$

$n = 6.25 \times 10^{23}$

Therefore, $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

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3 years ago

a glass bottle of soda is sealed with a screw cap the absolute pressure of the gas inside the bottle is 1.80*10^5 pa. assuming t

eduard

The force exerted by a pressure of any gas over a surface its given by the formula P=F/S (where P is pressure, F force and S surface).

We can multiply both sides of the formula by S to obtain the force.

P*S=(F*S)/S

P*S=F

Solve for P=1.80*10^5 Pa and S=4.10*10^-4 m^2 ([Pa] =[N/m^s])

(1.80*10^5 N/m^s) * (4.10*10^-4 m^2) =F

73.8 N =F

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3 years ago

The threatened or actual use of physical force to get someone to have sex without consent is called

Nimfa-mama [501]

Wouldn't it be r a p e?

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3 years ago

Read 2 more answers

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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3 years ago

Which property of gold allows it to be used this way?

densk [106]

the electric conductivity of gold is very high

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3 years ago

Please help me with my question ​

Please help me with my question