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Nutka1998 [239]
3 years ago
8

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of LaTe

X: 190~mm^2190 m m 2. How much charge must be transferred from one plate to the other if 1.1 nJ of energy are to be stored in the plates

Physics
1 answer:
RideAnS [48]3 years ago
7 0

Answer:

5.5x 10^-11 C

Explanation:

Pls see attached file

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In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the mea
Vikentia [17]
For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c). 

fringe = (delta t) / (λ/c) 

We can find (delta t) with the equation: 

delta t = [v^2(L1+L2)]/c^3 

Derivation of this formula can be found in your physics text book. From here we find (delta t): 

600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13 

2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes 

This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s. 
4 0
3 years ago
HURRY! PLEASE HELP!!!!<br><br><br> 3. What methods are you using to test this (or each) hypothesis?
Pie

Answer:

How to Test Hypotheses

State the hypotheses. Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis. ...

Formulate an analysis plan. The analysis plan describes how to use sample data to accept or reject the null hypothesis. ...

Analyze sample data. ...

Interpret the results.

4 0
3 years ago
A +0.05 C charge is placed in a uniform electric field pointing downward with a strength of 100 Newtons over Coulombs.. Determin
shtirl [24]

The magnitude of the electric force on the charge is 5 N.

<h3>Magnitude of force on the charge</h3>

The magnitude of force on the charge is calculated as follows;

F = Eq

where;

  • E is electric field
  • q is magnitude of the charge

F = 100 N/C  x 0.05 C

F = 5 N

Thus, the magnitude of the electric force on the charge is 5 N.

Learn more about electric force here: brainly.com/question/20880591

#SPJ1

8 0
2 years ago
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
3 years ago
Earth's gravitational pull just got threetimes stronger! what happens to your weight/
OLga [1]
Hold on and let's discuss this realistically.

Because of gravity, there are two forces between the Earth and me. One draws me toward the Earth. The strength of that force is what I call my "weight". The other force draws the Earth toward me, and has the same strength.

The strength of these forces depends on the masses of the Earth and me. If the strength just tripled, that means that at least one of us just picked up a lot more mass. If the Earth suddenly became three times as massive, then the weight of everything and everybody on it would suddenly triple, and I'm pretty sure it would be the end of all of us before too long.

If it was only MY mass that suddenly tripled, that would mean that I had gone tearing through my house and the neighbour's house, eating everything in sight including the 2 couches, 3 dogs, and 6 TVs. Naturally, just as you would expect, my weight changed from 207 to 621, and my skin is stretched really tight.
ooohhh
8 0
3 years ago
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