<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then:
F = G(Mm / (4(pi)*r^2))
The above expression gives the force that you feel on the earth's surface, as it is today!
Let us now double the mass of the earth and decrease its diameter to half its original size.
This is the same as replacing M with 2M and r with r/2.
Now the gravitational force (F' ) on the new earth's surface is given by:
F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F
So:
F' = 8F
This implies that the force that you would feel pulling you down (your weight) would increase by 800%!
You would be 8 times heavier on this "new" earth!</span>
Answer:
There would be more hours of sunlight at the equator
Transform boundary
is the answer from stemscopes
C) is correct
series circuit - in the same path : current flow on one path so they are equal on each component and equal to the source's. voltage on each components may be different.
parallel circuit - between same nodes : voltage of the components are equal and equal to the source's. current on each components may be different.
Answer: A
Explanation:Earthquakes occur on faults - strike-slip earthquakes occur on strike-slip faults, normal earthquakes occur on normal faults, and thrust earthquakes occur on thrust or reverse faults. When an earthquake occurs on one of these faults, the rock on one side of the fault slips with respect to the other.
