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UNO [17]
3 years ago
6

If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gr

avity at the surface would becomeA) 1/2 of what it now is.B) 2 times what it now is.C) 1/4 of what it now is.D) the same as it now is.E) 4 times what it now is.
Physics
1 answer:
Lerok [7]3 years ago
5 0

Answer:

B) 2 times what it now is

Explanation:

The acceleration due to gravity at the surface of the Earth is given by

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the mass of the Earth

R is the Earth's radius

In this problem, the mass of the Earth is doubled:

M' = 2M

while the radius remains the same:

R' = R

so the new acceleration due to gravity would be

g'=\frac{GM'}{R'^2}=\frac{G(2M)}{R^2}=2\frac{GM}{R^2}=2g

so, the acceleration due to gravity would become twice the current value.

Note also that the value of g does not depend on the mass of the objects involved.

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An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

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My answer -

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p.s

Glad to help you and if you need anything else on brainly let me know so I can elp you again have an AWESOME!!! :^)
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