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navik [9.2K]
2 years ago
7

Assume a planet is a uniform sphere of radius R that (somehow) has a narrow radial tunnel through its center. Also assume we can

position an apple anywhere along the tunnel or outside the sphere. Let FR be the magnitude of the gravitational force on the apple when it is located at the planet's surface.
a. How far from the surface is there a point where the magnitude is 1/2FR if we move the apple away from the planet?
b. How far from the surface is there a point where the magnitude is 1/2FR if we move the apple into the tunnel?
Physics
1 answer:
Drupady [299]2 years ago
5 0

Answer:

Explanation:

a )

Force at the surface = FR

Value of g at height h = g( 1 - 2 h / R )

force at height h = F_R ( 1 - 2 h / R)

F_R / 2 = F_R ( 1 - 2 h / R)

=  F_R   - 2  F_R xh / R

F_R / 2 = 2  F_R x h / R

1 / 2 = 2h / R

4h = R

h = R / 4 . Ans

b )

Value of g at depth  d = g( 1 -  d / R )

force at depth d  = F_R ( 1 - d / R)

F_R / 2 = F_R ( 1 - d / R)

=  F_R   -   F_R xd / R

F_R / 2 =   F_R x d / R

1 / 2 = d / R

2d  = R

d = R / 2

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Energy in a spring:

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k spring constant = 800 n/m
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Which of the following is a common downside of net pens, one of the most common forms of aquaculture?
Eduardwww [97]

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Answer is overcrowding aka answer choice A. I got the question and got it right. Please mark brainliest. Have a good day! :)

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denis-greek [22]

Answer:

No, it's not there.

Explanation:

For a machine to be 100% efficient, it has to be with an output which is equal to its input. But machines have an out put less than an input, hence efficiency below 100%.

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If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
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Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

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Substituting the values of the variables into the equation, we have

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Answer:

<em>3924 Pa</em>

<em></em>

Explanation:

Volume of cylinder = 2 L = 0.002 m^3  (1000 L = 1 m^3)

diameter of the inner cylinder = 8 cm = 0.08 m  (100 cm = 1 m)

radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m

area of the inner cylinder = \pi r^{2}

where \pi = 3.142,

and r = radius = 0.04 m

area of inner cylinder = 3.142 x 0.04^{2} = 0.005 m^2

<em>height h of the water in this cylinder = volume/area</em>

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<em>pressure at the bottom of the cylinder due to the height of water = pgh</em>

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g = acceleration due to gravity = 9.81 m/s^2

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