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saveliy_v [14]
3 years ago
9

How the relative motion and relative position of the Sun, Earth and Moon affect the seasons, phases of the moon and eclipses?

Physics
1 answer:
never [62]3 years ago
6 0
Maybe that seems me cold
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How many coulombs of charge do 50 * 10^31 electrons possess
Angelina_Jolie [31]
Quantity of Charge , Q = ne
Where n = number of electrons
             e = charge on one electron = -1.6 * 10 ^-19  C.
             n = 50 * 10^31  electrons

Q =    (50 * 10^31)*( -1.6 * 10 ^-19 ) =  -8 * 10^13 C.

Note that the minus sign indicates that the charge is a negative charge.
7 0
3 years ago
There are 2 concentric cylinders. These cylinders are very long with length L. The inner cylinder has a radius R1 and is a solid
OLga [1]
To get the charge along the inner cylinder, we use Gauss Law
E = d R1/2εo
For the outer cylinder the charge can be calculated using
E = d R2^2/2εoR1
where d is the charge density
Use these two equations to get the charge in between the cylinders and the capacitance between them.
5 0
3 years ago
Students perform an experiment in which they drop two eggs with equal mass from a balcony. In the first trial, the egg hits the
shepuryov [24]

<u> Answer </u>

The impulse on the second trial is smaller is smaller than in the first trial.

<u>Explanation </u>

Impose of a body is that change in momentum during a time interval. If the change of momentum takes longer then, the impulse of a force is less. I a moving object hits a hard surface the rate of change of momentum is very high. e.i in the first trial, the egg breaks because it hits the hard surface(ground).

In the second trial, the foam cushion absorbs the shock and prolongs the time of impact with the egg hence decreasing the impulse.


8 0
2 years ago
Read 2 more answers
Compare the signs of ƒ for lenses and mirrors.
STALIN [3.7K]

Answer:

simple

Explanation:

<h3>CONCAVE MIRRORS AND LENSES</h3>

<h3>f= negative</h3>

<h3>CONVEX MIRRORS AND LENSES</h3><h3 /><h3>f= positive</h3>

<h3>PLEASE FOLLOW ME AND MARK IT BRAINLIEST</h3>

3 0
2 years ago
Read 2 more answers
Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
3 years ago
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