Complete the following sentence. The World Wide Web launched to the public in.

In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe

sweet-ann [11.9K]

Answer:

Mass of the oil drop, $m=3.01\times 10^{-15}\ kg$

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

$qE=mg$

$m=\dfrac{qE}{g}$

$q=6e$ , e is the charge on electron

E is the electric field, $E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m$

$m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}$

$m=3.01\times 10^{-15}\ kg$

So, the mass of the oil drop is $3.01\times 10^{-15}\ kg$ . Hence, this is the required solution.

5 0

3 years ago

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

nydimaria [60]

Answer: a) the greater speed for the ball is getting with the large radius of the circle. b) 1.68* 10 ^3 m/s^2 c) 1.25*10^3 m/s^2

Explanation: In order to solve this problem firstly we have to consider that speed in a of the circular movement is directly the angular rotation multiply the radius of the circle so by this we found that the second radius get large speed.

Secondly to calculate the centripetal acceleration for the ball we have to considerer the relationship given by:

acceleration in a circular movement= ω^2*r

so

a1= (8.44 *2*π)^2*r1=1.68 *10^3 m/s^2

a2= (5.95*2*π)^2*r2=1.25*10^3 m/s^2

3 0

3 years ago

6/10

g100num [7]

Answer:pounds

Explanation:

7 0

2 years ago

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

atroni [7]

Answer:

The average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V$