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2 years ago

A negative oil droplet is held motionless in a millikan oil drop experiment. What happens if the switch is opened?

1 answer:

0 0

In Millikan oil drop experiment, when the switch is opened and by altering supply the charge of electron is determined.

Explanation:

Millikan's oil drop experiment is held to determine the terminal velocity and charge of the oil drop.

Firstly without any supply of voltage when an oil drop is sprinkled and these droplets gather electrons together and gives negative charge as they pass through air.

By applying and altering voltage applied on the plates, drop can be suspended in air. Millikan observed one drop after another, varying the voltage and noting the effect. After many repetitions he concluded that charge could assume only certain fixed values.

After conducting many times he concluded 1.602176487 ×10−19 C as the charge of an electron.

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A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r

Firlakuza [10]

Answer:

Speed of the this part is given as

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

$P_1 + P_2 + P_3 = P_i$

here we know that

$P_1 = P_2$

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

$P_1 + P_2 = \sqrt{P^2 + P^2}$

$P_1 + P_2 = \sqrt2 mv$

now from above equation we have

$P_3 = -(P_1 + P_2)$

$mv_3 = -(\sqrt 2 mv)$

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

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2 years ago

Pls hurry I’m timed Which describes the formation of horizon B?

ioda

First one but I am not sure

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2 years ago

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Magma usually stays below the Earth's

Vesna [10]

Answer:

In the inner vote part of the earth

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2 years ago

If two objects have different temperatures when they come in contact, heat will flow from the warmer

r-ruslan [8.4K]

Answer:

They both have an equal temperature.

Explanation:

Any material or object that allow the conduction (transfer) of electric charge or thermal energy is generally referred to as a conductor. Some examples of a conductor are metals, copper, aluminum, graphite, etc.

In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.

Hence, if two objects have different temperatures when they come in contact, heat will flow from the warmer object to the cooler one until they both have an equal temperature.

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2 years ago

coin 1 is thrown upward from the top of 100m tower with a speed of 15m/s. coin 2 is dropped from the top of the tower 2.0second

kvv77 [185]

The height below the tower at which coin 1 pass coin 2 is 89.04 m.

The given parameters:

height of the tower, h = 100 m

initial velocity of coin 1, v = 15 m/s

time spent in air by coin 1 before coin 2 was dropped = 2s

To find:

the height below the tower at which coin 1 passes coin 2

Find the maximum height attained by coin 1 before falling to the ground:

$v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m$

Find the time taken for coin 1 to fall to the ground:

Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m

$t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s$

But the time taken for the coin 1 to reach 11.25 m above the tower:

$t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s$

Total time spent by coin 1 before reaching ground with respect to coin 2:

time = (1.5 s + 4.72 s) - 2 s

time = 4.22 s

<u>Note:</u><em> the 2 s was subtracted to keep both coins at a fair starting time below the tower</em>.

Find the total time taken for coin 2 to fall to the ground:

Height of coin 2 above the ground = 100 m

Total time taken by coin 2 before falling to the ground is calculated as:

$t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s$

The time at which coin 1 will pass coin 2 is 4.22 s.

Find the height below the tower when the time is 4.22 s.

$h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m$

Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.

Learn more here: brainly.com/question/18943273

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3 years ago