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Assoli18 [71]
2 years ago
10

How did Millikan's oil drop experiment lead to quantum nature of electric charge?​

Physics
1 answer:
ahrayia [7]2 years ago
5 0

Answer:

Millikan's oil drop experiment proved that electric charge is quantized.

At the time, there was still great debate whether electric charge was continuous or not.

Millikan believed that there was a smallest unit of charge, and he set out to prove it.

This was the big result of the oil drop experiment.

That he could also determine the charge of the electron was a secondary benefit.

This was probably one of the most significant experiments ever carried out.

Explanation:

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Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her
leva [86]

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

\tau_l = F*d

And in angular movement like

\tau_a = I*\alpha

Where,

F= Force

d= Distance

I = Inertia

\alpha = Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

\tau= F*cos(19)*d

On the other hand we have the speed data expressed in RPM, as well

\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 1.0471rad/s

Acceleration can be calculated by

\alpha = \frac{\omega_f}{t}

\alpha = \frac{1.0471}{9}

\alpha = 0.11rad/s^2

In the case of Inertia we know that it is equivalent to

I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2

I = 1983.75kg.m^2

Matching the two types of torque we have to,

\tau_l=\tau_a

Fd=I\alpha

Fcos(19)*2.3=1983.75(0.11)

F=100.34N

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

W = \frac{1}{2}I\omega_f^2

W= \frac{1}{2}(1983.75)(1.0471)^2

W=1087.51J

7 0
2 years ago
A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet
Leno4ka [110]

Answer:

0.94 m³/s

Explanation:

From the question given above, the following data were obtained:

Air flow (in ft³/min) = 2×10³ ft³/min

Air flow (in m³/s) =.?

Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:

35.315 ft³/min = 1 m³/min

Therefore,

2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min

2×10³ ft³/min = 56.63 m³/min

Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:

1 m³/min = 1/60 m³/s

Therefore,

56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min

56.63 m³/min = 0.94 m³/s

Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.

8 0
3 years ago
A 55 meter rope is lying on the floor and has has a weight of 4040 N in total. How much work is required to lift up one end of t
Kazeer [188]

Answer:

Explanation:

Given

Length of rope L=55\ m

Weight of rope W=4040\ N

weight density \lambda =\frac{4040}{55}=73.45\ N/m

Work done to lift rope 33 m

W=\int_{0}^{33}\lambda hdh

W=\int_{0}^{33}73.45hdh

W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0

W=39.993\ kJ      

6 0
3 years ago
What is work? how do we solve for work?​
slava [35]

Answer: Work can be calculated with the equation: Work = Force × Distance. The SI unit for work is the joule (J), or Newton • meter (N • m). One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m.

Explanation:

5 0
2 years ago
Read 2 more answers
An embryo develops inside the ________.​
irina1246 [14]
Development of the Embryo. The next stage in development is the embryo, which develops within the amniotic sac, under the lining of the uterus on one side. This stage is characterized by the formation of most internal organs and external body structures.
6 0
3 years ago
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