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1 year 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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Step-by-step explanation:

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1 year ago

An auto moves 10 meters in the first second of travel, 15 more meters in the next second, and 20 more meters during the third se

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1 year ago

A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp $h_{1}=2.30\ m$

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Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

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$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

We need to calculate the velocity of the ball

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

Using conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

Put the value into the formula

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

$d =vt$

Where. d = distance

t = time

v = velocity

Put the value into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

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1 year ago

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The answer is True. The amount force exerted by any object is directly proportional to its mass. This means that our planet is exerting more gravitational force to Angelina, and Angelina is also exerting a gravitational force on our planet directly proportional to her mass. Angelina is actually falling towards the center of the earth,and also our planet is also moving towards Angelina, but it seems negligible with respect to Angelina.Our Sun is so massive that it held our planet in its orbit because of its gravitational force.

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B <<<<====== answer.

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