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maria [59]
3 years ago
15

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 ft. thick) hollow spherical shell below your feet. Also suppose there was a small hole in the shell, just big enough four you to fit through. Compare the force of gravity on you outside the shell to the force of gravity if you stepped inside the shell.
Physics
1 answer:
KiRa [710]3 years ago
3 0

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>

<em></em>

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/R^{2}

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/1^{2}

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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Answer:

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Explanation:

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Answer:

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3 years ago
How long will it take for a boat traveling at 36 k/h to go 126 km?
brilliants [131]
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rs. Rushing fills a balloon with hydrogen gas to demonstrate its ability to burn. Which combination could she use to produce the
anyanavicka [17]

Answer:

Mg and HCl

Explanation:

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Read 2 more answers
A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air
fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

s=u_y t+\frac{1}{2}at^2

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

u_y=0 is the initial vertical velocity

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Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

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Therefore, the horizontal distance covered after a time t is

d=v_x t

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

d=(10)(2.02)=20 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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