Number 1 is D number 2 is C
Answer:
Charge on Moon and Earth is 5.43x10¹³ C .
Explanation:
Gravitational force is the force of attraction between any two bodies having mass while Electrostatic force is the force experienced by two charge bodies. Electrostatic force can be attractive or repulsive.
Let M and m be the mass of Earth and Moon respectively, d is the distance between Earth and Moon and q be the charge on Earth and -q be on the Moon.
Gravitational force, F₁ =![\frac{G\times{m}\times{M}}{d^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BG%5Ctimes%7Bm%7D%5Ctimes%7BM%7D%7D%7Bd%5E%7B2%7D%20%7D)
Here G is gravitational constant.
Electrostatic force, F₂ = ![\frac{k\times{q}\times{q}}{d^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7Bk%5Ctimes%7Bq%7D%5Ctimes%7Bq%7D%7D%7Bd%5E%7B2%7D%20%7D)
Here k is Coulomb constant.
According to the problem, the gravitational force between Earth and Moon is equal to the electrostatic force between them.
F₁ = F₂
= ![\frac{k\times{q}\times{q}}{d^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7Bk%5Ctimes%7Bq%7D%5Ctimes%7Bq%7D%7D%7Bd%5E%7B2%7D%20%7D)
= ![q^{2}](https://tex.z-dn.net/?f=q%5E%7B2%7D)
Put 6.07x10⁻¹¹ N m²/kg² for G, 5.97x10²⁴ kg for M, 7.34x10²² kg for m and 9x10⁹ N m²/C² in the above equation.
= q²
q = ![\sqrt{2.95\times{10^{27} }}](https://tex.z-dn.net/?f=%5Csqrt%7B2.95%5Ctimes%7B10%5E%7B27%7D%20%7D%7D)
q = 5.43x10¹³ C
Answer is
9.773m/s^2
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Given,
h=8848m
The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.
g′=g(1 − 2h/h)
=9.8(1 - 6400000/17696)
=9.8(1 − 0.00276)
9.8×0.99724
=9.773m/s^2
Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2
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hope this helps :)
Answer:
The magnitude of the electric field inside the membrane is 8.8×10⁶V/m
Explanation:
The electric field due to electric potential at a distance Δs is given by
E=ΔV/Δs
We have to find the magnitude electric field in the membrane
Ecell= -ΔV/Δs
![E_{cell}=-\frac{V_{in}-V_{out}}{s} \\E_{cell}=\frac{V_{out}-V_{in}}{s}\\E_{cell}=\frac{0.070V}{8*10^{-9}m } \\E_{cell}=8.8*10^{6}V/m](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D-%5Cfrac%7BV_%7Bin%7D-V_%7Bout%7D%7D%7Bs%7D%20%5C%5CE_%7Bcell%7D%3D%5Cfrac%7BV_%7Bout%7D-V_%7Bin%7D%7D%7Bs%7D%5C%5CE_%7Bcell%7D%3D%5Cfrac%7B0.070V%7D%7B8%2A10%5E%7B-9%7Dm%20%7D%20%5C%5CE_%7Bcell%7D%3D8.8%2A10%5E%7B6%7DV%2Fm)
The magnitude of the electric field inside the membrane is 8.8×10⁶V/m
Answer:
i have absolutly no idea how to do it but i looked it up and your answer should be B. i could be wrong but thats what the web told me