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Juli2301 [7.4K]
3 years ago
7

If Earth had twice the mass it has now, how would the gravitational force between it and the Sun change?

Physics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

a)The gravitational force would be twice as much as it is now

Explanation:

If the earth had twice its mass as it is now, the gravitational force between the sun and the earth would be twice as much as it is now.

According to Newton's law of universal gravitation "the gravitational force of attraction between bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them".

       F_{g}   = \frac{G m _{e} m _{s}  }{r^{2} }

G is the universal gravitation constant

m is the mass

r is the distance

let mass of sun  = s

     mass of earth  = e

 new mass of sun  = s

 new mass of sun  = 2e

input variables;

    F  = \frac{Gx e x s}{r^{2} }    ------ i

   F = \frac{G  x s x 2e}{r^{2} }    ------- ii

From the second equation;

      2F  = \frac{Gx e x s}{r^{2} }

      2F  = F

Therefore, the force will double.

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Help help help please , i will give brainliest
finlep [7]

Answer:

I believe the answer for #1 is D and the answer for #2 is B

Explanation:

I hope this is correct and helps

6 0
3 years ago
Read 2 more answers
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0
3 years ago
What are some of the downsides of using hydroelectric power?.
never [62]
Has an Environmental Impact. Perhaps the largest disadvantage of hydroelectric energy is the impact it can have on the environment.
It Displaces People.
It's Expensive.
There are Limited Reservoirs.
There are Droughts.
It's Not Always Safe
4 0
2 years ago
a bird flies 25.0 m in the direction 55° east of south to its nest. the bird then flies 75.0 m in the direction 55° west of nort
son4ous [18]

The northward components of the resultant displacement is 40.96 m and the westward components of the resultant displacement of the bird from its nest is 28.68 m.

<h3>Displacement of the bird</h3>

The displacement of the bird is the change in the position of the bird.

<h3>Vertical component of the bird's displacement </h3>

Vy₁ = -25 m  x   sin(55)

Vy₁ = -20.48 m

Vy₂ = 75 m   x    sin(55)

Vy₂ = 61.44 m

Total vertical displacement = 61.44 m - 20.48 m = 40.96 m

<h3>Horizontal component of the bird's displacement </h3>

Vx₁ = -25 m  x   cos(55)

Vx₁ = -14.34 m

Vx₂ = 75 m   x    cos(55)

Vx₂ = 43.02 m

Total horizontal displacement = 43.02 m - 14.34 m = 28.68 m

Learn more about displacement here: brainly.com/question/2109763

#SPJ1

8 0
1 year ago
Explain the relationship between the earth's crust and the earth's ocean sizes.
valina [46]

the outermost layer of Earth’s lithosphere that is found under the oceans and molded at scattering centres ono ceanic ridges, which occur at deviating plate boundaries

Oceanic crust is about 6 km (4 miles) thick.

hope it helps

3 0
3 years ago
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