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3 years ago

When the distance between two stars decreases by one-third, the force between them

Physics

1 answer:

pashok25 [27]3 years ago

4 0

Answer:

the force will increase by a factor 2.25

Explanation:

The gravitational force between the two stars is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two stars

r is the distance between the stars

If the distance is decreased by one-third, it means that the new distance is 2/3 of the previous distance

$r'=\frac{2}{3}r$

So the new force will be

$F'=G\frac{m_1 m_2}{(\frac{2}{3}r)^2}=\frac{9}{4} G\frac{m_1 m_2}{r^2}=2.25 F$

So, the force will be 2.25 times the previous value.

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Answer:

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Explanation:

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3 years ago

A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a

asambeis [7]

Answer:

The recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

$m=\frac{F}{g}$

Calculate the combined mass of the man and rifle.

$m_{man,rifle}=\frac{650+25}{g}$

Put $g=9.8 ms^{-2}$ .

$m_{man,rifle}=\frac{650+25}{9.8}$

$m_{man,rifle}=68.88 kg$

The expression for the conservation of momentum is as follows as;

$m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}$

Here, $m_{man,rifle}$ is the mass of the man and rifle, $m_{rifle}$ is the mass of the rifle, $u_{man},u{bullet}$ are the initial velocities of the man and bullet and $v_{man},v{man,rifle}$ are the final velocities of the man and rifle and rifle.

It is given in the problem that a rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s.

Convert mass of rifle from gram to kilogram.

$m_{bullet}=4.5 g$

$m_{bullet}=.0045 kg$

Put $m_{bullet}=.0045 kg$ , $m_{man,rifle}=68.88 kg$ , $u_{man,rifle}=0$ , $v_{bullet}= 240 ms^{-1}$ and $u_{bullet}=0$ .

$m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+1.08$

$(68.88)v_{man,rifle}=\frac{-1.08}{68.88}$

$v_{man,rifle}=-0.016 ms^{-1}$

Therefore, the recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

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3 years ago

Explain the importance of measurement<br>Physics.<br><br>

KiRa [710]

Answer:

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3 years ago

The study of charges in motion and their

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Answer:

The study of charges in motion and their interaction with magnetic fields is known as electromagnetism.

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3 years ago

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

ankoles [38]

<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3 s

Substituting

s = ut + 0.5 at²

s = 0 x 3 + 0.5 x 9.81 x 3²

s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

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3 years ago