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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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ACUTE INFECTION plz help worth 70 pt plz help !!!!! Also brainlest
Paraphin [41]

Answer:

Simple awnser Do it yourself I really would help but I have no clue! Sorry

Explanation:

3 0
3 years ago
(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg
valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be m_ 1 = 80 kg

The weight of the buddy be m_2 = 120 kg

The weight of  Bubba be m_3 = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2

The length of the boat be x_2 = 4 m

∴

We can find the center of mass of the system by using the formula:

X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}

4 0
3 years ago
A uniform brick of length 21 m is placed over
Paha777 [63]

Answer:

15.75 m

Explanation:

First, let's look at the top brick by itself.  In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick.  So the edge of the top brick must be 10.5 m from the edge of the bottom brick.

Now let's look at both bricks as a combined mass.  We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m.  And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge.  So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.

6 0
4 years ago
A small object of mass 3.82 g and charge -16.5 µC is suspended motionless above the ground when immersed in a uniform electric f
horrorfan [7]

Answer:

2271.16N/C  upward

Explanation:

The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.

First we determine the downward weight using

W=mg\\g=9.81m/s^{2}

Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be

W=0.00382kg*9.81m/s^{2}

W=0.0375N

To calculate the electric field,

E=f/q\\E=0.0375/16.5*10^{-6} \\E=2271.16N/C

Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is  repelling it, Hebce we can conclude that the electric field lines are upward.

Hence the magnitude of the electric force is 2271.16N/C and the direction is upward

4 0
3 years ago
An 800-g block of ice at 0.00°C is resting in a large bath of water at 0.00°C insulated from the environment. After an entropy c
Allisa [31]

Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

          ΔH is enthalpy change

          T is absolute temperature

          ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

           Entropy change,ΔS = 600 J/K

           Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

  ∴ΔH = 273 x 600

           = 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

  ∴ΔH = mass of ice melted x latent heat of fusion of water

    Mass of ice melted = ΔH / latent heat of fusion of water

                                     = 163800 / 333

                                     = 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

                                                   = 800-491.891

                                                  = 308.109 g

Amount of ice remained after melting = 308.109 g

8 0
3 years ago
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