In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass.
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were.
Ok, Rameses:
Elementary charge . . . . . 1.6 x 10⁻¹⁹ coulomb
negative on the electron
plussitive on the proton
Electron rest-mass . . . . . 9.11 x 10⁻³¹ kg
a). The force between two charges is
F = (9 x 10⁹) Q₁ Q₂ / R²
= (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²
= ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²
= 8.05 x 10⁻⁸ Newton .
b). Centripetal acceleration =
v² / r .
A = (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)
= 7.7 x 10²² m/s² .
That's an enormous acceleration ... about 7.85 x 10²¹ G's !
More than enough to cause the poor electron to lose its lunch.
It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use " F = MA "
either.
I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether F = M A .
I'm going to leave that step for you. Good luck !
Answer:

Explanation:
Given that,
Angular speed of a skater, 
The moment of inertia of the skater, I = 0.6 kg-m²
We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

Substitute all the values,

So, its angular momentum is equal to
.
Answer:
<em>The first law states that</em> every planet describes an elliptical path about the sun as a single focus.
<em>The</em><em> </em><em>second</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The line joining the planet to the sun sweeps out equal areas in equal time intervals.
<em>The</em><em> </em><em>third</em><em> </em><em>law</em><em> </em><em>states</em><em> </em><em>that</em><em> </em>The squares of the period of revolution is proportional to the cubes of the mean distance between the planet and the sun
The answer is volt for this question.
I hope is correct if not I’m sorry.