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3 years ago

Newton's Law of Gravity states that if the mass of one object doubles, the gravitational pull on a second object will

Physics

2 answers:

Yakvenalex [24]3 years ago

8 0

Answer:

It doubles.

If the mass of one of the objects is doubled, then the force of gravity between them is doubled. Since the gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

OLEGan [10]3 years ago

5 0

Answer:

It will double.

Explanation:

Newton's Law of Gravity states that $F_G=\frac{Gm_1m_2}{r^{2}}$ , where G is the gravitational constant, r is the distance between the objects' centers, and $m_1$ and $m_2$ are the objects' masses. We just have simple math here: by doubling $m_1$ , we double the entire fraction. By doubling the entire fraction, we double the gravitational pull. Therefore, Newton's Law of Gravity states that if the mass of one object doubles, the gravitational pull on a second object will double.

I hope this helps you understand it! Have a great day, 'kay?

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while working out a man performed 2525j of work in 19seconds . what was his power A:132.9w. B:241.5w C 47.975w. D100.5w

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The inductance of a closely packed coil of 560 turns is 8.9 mH. Calculate the magnetic flux through the coil when the current is

Y_Kistochka [10]

Answer:

$0.11\times 10^{-6}weber$

Explanation:

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svetoff [14.1K]

Answer:

Explanation:

There are two hypotheses she could test:

A cat's heart rate changes while it is napping.

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A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

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mg = Weight of object = 20 N

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v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

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