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 !
I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
Assuming acceleration due to gravity of the moon is constant and there’s no initial velocity in the mans jump we can use one of the kinematic equations. x(final)=x(initial)+(1/2)gt^2. Plug in known values. 0=10-(1.62/2)t^2. The value 1.62 is acceleration of gravity on the moon. Now simply solve for t. t=3.513
Sorry I don’t know the exact answer