In part A of the lab we see that the magnetic field of a long straight wire __. a. increases with distance in a linear relations
1 answer:
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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.
Answer:
Explanation:
<u>Given:</u>
No. of times = 600 times
Time = 60 seconds
<u>Required:</u>
Frequency = f = ?
Time Period = T = ?
<u>Solution:</u>
<h2>Frequency = No. of Times / Time</h2>Frequency = 600 / 60
<u>Frequency = 10 Hz</u>
<h2>Time Period = 1 / Frequency</h2>Time Period = 1 / 10
<u>Time Period = 0.1 secs</u>