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2 years ago

What is it called when air particles spread apart?

1 answer:

5 0

When air particles bunch up, it's called a compression-rarefaction.
When the particles spread apart its called a rarefaction.

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An electron is accelerated through a potential difference of 3.1 kV and directed into a region between two parallel plates separ

Lapatulllka [165]

Answer: 178.25*10^-6 T

Explanation: In order to solve this problem we have to take into account the equilibrium between the electric and magnetic forces in the electron, so it is given by:

Fm=evB

Fe=eE so

evB=eE the we have

v=E/B

Firsly we calculate the velocity of the electron before to get the parallel plates at 100V

eΔV=1/2*m*v^2 then

v=(2*eΔV/m)^1/2

v=(2*1.6*10^-19*3.1*10^3/9.1*10^-31)^1/2=33 *10^6 m/s

Then we can calculate B

B=E/v E.d=V where d is the separation between the plates and V is equal a 100V

B=V/(d*v)=100/(17*10^-3*33 *10^6)=178.25*10^-6 T

3 0

3 years ago

Use each of the following terms in a separate

Alenkinab [10]

I think the correct answer is going to be A

3 0

2 years ago

In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single

FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is $2.27\times 10^{39}$

Explanation:

It is given that,

Distance between electron and proton, $r=4.53\ A=4.53\times 10^{-10}\ m$

Electric force is given by :

$F_e=k\dfrac{q_1q_2}{r^2}$

Gravitational force is given by :

$F_g=G\dfrac{m_1m_2}{r^2}$

Where

$m_1$ is mass of electron, $m_1=9.1\times 10^{-31}\ kg$

$m_2$ is mass of proton, $m_2=1.67\times 10^{-27}\ kg$

$q_1$ is charge on electron, $q_1=-1.6\times 10^{-19}\ kg$

$q_2$ is charge on proton, $q_2=1.6\times 10^{-19}\ kg$

$\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}$

$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}$

$\dfrac{F_e}{F_g}=2.27\times 10^{39}$

So, the ratio of electric force to the gravitational force is $2.27\times 10^{39}$ . Hence, this is the required solution.

3 0

2 years ago

You need to put a metal rod

Lubov Fominskaja [6]

-- Put the rod into the freezer for a while. As it cools,
it contracts (gets smaller) slightly.

-- Put the cylinder into hot hot water for a while. As it heats,
it expands (gets bigger) slightly.

-- Bring the rod and the cylinder togther quickly, before the
rod has a chance to warm up or the cylinder has a chance
to cool off.

-- I bet it'll fit now.

-- But be careful . . . get the rod exactly where you want it as fast
as you can. Once both pieces come back to the same temperature,
and the rod expands a little and the cylinder contracts a little, the fit
will be so tight that you'll probably never get them apart again, or even
move the rod.

8 0

2 years ago

If object A has more mass than object B, what will object A need to accelerate at the same rate as object B?

Leni [432]

Answer:

More force

Explanation:

Object A has more mass than object B

For object A to accelerate at the same rate as object B, it will need more force.

According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".

Net force = mass x acceleration

Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.

3 0

2 years ago