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lara [203]
3 years ago
9

Gravitational attraction depends on the mass of the objects as well as their distance. The gravitational force between objects i

ncreases as the masses of the objects increase. The gravitational force decreases as the distance between objects increases. The impact of mass and distance are not equal. If the mass of one object doubles, the gravitational force also doubles. If the distance between objects doubles, the gravitational force decreases by 1/4. Using the model predict what will happen to the gravitational force if the masses both double and the distance also doubles.
Physics
2 answers:
shepuryov [24]3 years ago
7 0
<h2>Answer: Gravitational attraction will be the same</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}    (1)

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

Now, if we double both masses and the distance also doubles, this means:

m_{1} and m_{2} will be now 2m_{1} and 2m_{2}

r will be now 2r

Let's rewrite the equation (1) with this new values:

F=G\frac{(2m_{1})(2m_{2})}{(2r)^2}    (2)

Solving and simplifying:

F=4G\frac{m_{1}2m_{2}}{4r^2}    

F=G\frac{m_{1}m_{2}}{r^2}     (3)

As we can see, equation (3) is the same as equation (1).

So, if the masses both double and the distance also doubles the <u>Gravitational attraction between both masses will remain the same.</u>

Dmitry [639]3 years ago
7 0

Answer:

C) The gravitational attraction remains constant.

even though the distance is double The gravitational attraction remains constant no matter what.

I hope this helps!!!

Your welcome,

                         Shadow Lolbit

Can we be friends!!!!

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3 years ago
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
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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.

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3 years ago
The Earth’s internal __________ source provides the energy for our dynamic planet, providing it with the driving force for on-go
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A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They
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Answer:

a.\thta=71^{\circ}

b.v_f=3.78 m/s

Explanation:

We are given that

m_1=53 kg

v_1=2.9 m/s

m_2= 72 kg

v_2=6.2 m/s

a.We have to find the angle

\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}

\theta=71^{\circ}

b. We have to find the speed v_f

According to law of conservation of momentum

m_1v_1=(m_1+m_2)v_fcos\theta

53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f

v_f=\frac{153.7}{40.7}=3.78 m/s

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The average speed between 0 h and 2.340 h is 6.97 Km/h

Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.

Average \: speed =  \frac{total \: distance}{total \: time}  \\  \\

With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:

  • Total time = 2.340 – 0 = 2.340 h
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Average \: speed =  \frac{total \: distance}{total \: time}  \\  \\Average \: speed =  \frac{16.3}{2.340}  \\  \\ Average \: speed = 6.97 \: Km/h \\  \\

Learn more about average speed: brainly.com/question/24884027

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