Now suppose that the Earth had the same mass and radius as it currently does, but all of the mass was concentrated into a thin (
1 answer:
Answer:
The force of gravity at the shell will be extremely great on me due to the huge mass collapsed into the small radius.
<em>At the center of the shell, the gravitational forces all around should cancel out, giving me a feeling of weightlessness; which will be a lesser force compared to that felt while standing on the shell.</em>
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Explanation:
For the collapsed earth:
mass = 5.972 × 10^24 kg
radius = 1 ft
according to Newton's gravitation law, the force of gravity due to two body with mass is given as
Fg = GMm/
Where Fg is the gravitational force between the two bodies.
G is the gravitational constant
M is the mass of the earth
m is my own mass
R is the distance between me and the center of the earths in each case
For the case where I stand on the shell:
radius R will be 1 ft
Fg = GMm/
Fg = GMm
For the case where I stand stand inside the shell, lets say I'm positioned at the center of the shell. The force of gravity due to my mass will be balanced out by all other masses around due to the shell of the hollow earth. This cancelling will produce a weightless feeling on me.
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The electric field at one corner of a square is 1614217 N/C.
Explanation:
The distance between x and y direction diagonals.
As per the given details the distance between diagonals is calculated as
0.5² + 0.5² = c² => c = 0.707 m
Charge to the right: In x direction
In order to find the electric charge towards x direction
we use e = kq/r² formula
As 'k' is coulomb's constant it's value is 9 x N m²/C²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.707)²] cos 45
e = 225000√2 N/C
X direction sum = 1218198 N/C.
Similarly as shown in x direction the charge is same for y direction also
Charge below: For y direction
e = kq/r²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.5)²] sin 45
e = 159099 N/C
Y direction sum = 1059099 N/C
Resultant electric field strength:
1218198 ² + 1059099² = e²
e = 1614217 N/C [45 degrees below the horizontal]
Answer:
108 km
Explanation:
The conversion factor between meters and feet is
1 m = 3.28 ft
So the second altitude, written in feet, can be rewritten in meters as
or in kilometers,
the first altitude in kilometers is
so the difference between the two altitudes is