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ss7ja [257]
1 year ago
9

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

Physics
1 answer:
scoray [572]1 year ago
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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Consider two slides, both of the same height. One is long and the other is short. From which slide will a child have a greater f
Lunna [17]

Answer:

The final speed will be the same for the children on the shorter side and on the longer side.

Explanation:

This is because since the they are the same distance above the ground, their potential energy which is a function of mass, acceleration due to gravity and vertical height are the same.

PE = Mass × gravity × vertical height

At this point, we can deduce that the horizontal length of the slide has no effect on the potential energy. Only the vertical height does.

All this potential energy is converted to kinetic energy at the end of the slide. Since the potential energy is the same, then the kinetic energy will be the same and thus their velocity is the same.

Mathematically, consider that PE = mgh and KE = \frac{1}{2}mv^{2}

at the bottom of the slide, since energy has to be conserved, PE must be equal to KE.

mgh = \frac{1}{2}mv^{2}

final velocity of the child , v = \sqrt{2gh}

It shows the final velocity is only a function f acceleration due to gravity and height.

Thus, making their velocities equal.

8 0
1 year ago
Rachel has an unknown sample of a radioisotope listed in the table. Using a special technique, she is able to measure the mass o
mart [117]
<span>Barium-139 is the correct answer.</span>
3 0
2 years ago
Read 2 more answers
If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric fiel
Grace [21]

Answer:

There will be no change in the direction of the electric field .

Explanation:

The direction will remain the same because the sign of the charges has no effect on it.

When one replaces the conducting cube with one that has positive charge carriers there will be no change in the direction of the field as there is no defined relationship between the direction of the electric field and sign of the charge.

3 0
2 years ago
Julius competes in the hammer throw event. The hammer has a mass of 7.26 kg and is 1.215 m long. What is the centripetal force o
nevsk [136]
In the circular motion of the hammer, the centripetal force is given by
F=m \frac{v^2}{r}
where m is the mass of the hammer, v its tangential speed and r is the distance from the center of the motion, i.e. the length of the hammer.
Using the data of the problem, we find:
F=m \frac{v^2}{r}=(7.26 kg) \frac{(31.95 m/s)^2}{1.215 m}=6100 N
4 0
2 years ago
Read 2 more answers
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=\nu=1.2Hz

Total energy of the oscillation=0.51 J

We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Hence, the amplitude of oscillation=A=0.199

4 0
1 year ago
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