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2 years ago

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Physics

2 answers:

TiliK225 [7]2 years ago

6 0

Answer:

C: An increase in the distance between the objects causes a greater change in the gravitational force than the same increase in mass.

Explanation:

it's C on edge! hope this helps!!~ (❁´▽`❁)*✲ﾟ*

stellarik [79]2 years ago

5 0

Answer:

(C) an increase in tue distance between the ibject causes a greater change in the gravitational force than the same increase in mass

Hope this helps

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Two identical stars with mass M orbit around their center of mass. Each orbit is circular and has radius

kondaur [170]

Answers: (A) $F=G\frac{M^2}{4R^2}$ (B) $V=\sqrt{\frac{GM}{4R}}$ (C) $T=4\pi R\sqrt{\frac{R}{GM}}$ (D)

$E=-\frac{GM^{2}}{4R}$

Explanation:

<h2>(A) Gravitational force of one star on the other</h2>

According to the law of universal gravitation:

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

Where:

$F$ is the module of the gravitational 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

In the case of this binary system with two stars with the same mass $M$ and separated each other by a distance $2R$ , the gravitational force is:

$F=G\frac{(M)(M)}{(2R)^2}$ (2)

$F=G\frac{M^2}{4R^2}$ (3) This is the gravitational force between the two stars.

<h2>(B) Orbital speed of each star</h2>

Taking into account both stars describe a circular orbit and the fact this is a symmetrical system, the orbital speed $V$ of each star is the same. In addition, if we assume this system is in equilibrium, <u>gravitational force must be equal to the centripetal force</u> $F_{C}$ (remembering we are talking about a circular orbit):

So: $F=F_{C}$ (4)

Where $F_{C}=Ma_{C}$ (5) Being $a_{C}$ the centripetal acceleration

On the other hand, we know there is a relation between $a_{C}$ and the velocity $V$ :

$a_{C}=\frac{V^{2}}{R}$ (6)

Substituting (6) in (5):

$F_{C}=M\frac{V^{2}}{R}$ (7)

Substituting (3) and (7) in (4):

$G\frac{M^2}{4R^2}=M\frac{V^{2}}{R}$ (8)

Finding $V$ :

$V=\sqrt{\frac{GM}{4R}}$ (9) This is the orbital speed of each star

<h2>(C) Period of the orbit of each star</h2><h2 />

The period $T$ of each star is given by:

$T=\frac{2\pi R}{V}$ (10)

Substituting (9) in (10):

$T=\frac{2\pi R}{\sqrt{\frac{GM}{4R}}}$ (11)

Solving and simplifying:

$T=4\pi R\sqrt{\frac{R}{GM}}$ (12) This is the orbital period of each star.

<h2>(D) Energy required to separate the two stars to infinity</h2>

The gravitational potential energy $U_{g}$ is given by:

$U_{g}=-\frac{Gm_{1}m_{2}}{r}$ (13)

Taking into account this energy is always negative, which means the maximum value it can take is 0 (this happens when the masses are infinitely far away); the variation in the potential energy $\Delta U_{g}$ for this case is:

$\Delta U_{g}=U-U_{\infty}$ (14)

Knowing $U_{\infty}=0$ the total potential energy is $U$ and in the case of this binary system is:

$U=-\frac{G(M)(M)}{2R}=-\frac{GM^{2}}{2R}$ (15)

Now, we already have the <u>potential energy</u>, but we need to know the kinetic energy $K$ in order to obtain the total <u>Mechanical Energy</u> $E$ required to separate the two stars to infinity.

In this sense:

$E=U+K$ (16)

Where the kinetic energy of both stars is:

$K=\frac{1}{2}MV^{2}+\frac{1}{2}MV^{2}=MV^{2}$ (17)

Substituting the value of $V$ found in (9):

$K=M(\sqrt{\frac{GM}{4R}})^{2}$ (17)

$K=\frac{1}{4}\frac{GM^{2}}{R}$ (18)

Substituting (15) and (18) in (16):

$E=-\frac{GM^{2}}{2R}+\frac{1}{4}\frac{GM^{2}}{R}$ (19)

$E=-\frac{GM^{2}}{4R}$ (20) This is the energy required to separate the two stars to infinity.

4 0

3 years ago

A light spring with a force coefficient 11.85 N/m is compressed by 14 cm as it is held between a 0.27 kg block on the left and a

lilavasa [31]

Answer:

A) Left = 6.14 m/s2 Right=2.765 m/s2 B) Left = 4.59 m/s2 Right= 1.215 m/s2

Explanation:

In the question we are given the spring constant which is 11.85 N/m and the compression of the spring is 14 cm. There are two blocks in front of the springs which are 0.27 kg and 0.6 kg respectively.

First, we need to calculate how much force the spring are going to exert on the masses when they are released. Since this force is not going to change for both cases, we only need to do it once.

The formula for calculation the force is F = k.x where k is the spring constant or force coefficient 11.85 N/m and the x is the compression which is 14 cm or 0.14 meters. If we put them in the equation we can find that F = 1.659 N

A) In the first case scenario, where the friction is equal to 0, we can use the formula F=m.a where F is the force applied by the release of the spring, m is the mass of the block and a is the acceleration.

For the first block, when we put 1.659 N and 0.27kg in the equation, a is calculated to be 6.14 m/s2.

For the second block, the same force of 1.659 N and 0.6 kg, a is calculated to be 2.765 m/s2.

B) In the second case scenario, where the friction is equal to 0.158, we first need to calculate its effect on each block. We need to use the formula Fk = μ.N where μ is the friction constant and N is the normal force of the block which is m.g (where g = 9,81 m/s2).

The friction force for the first block is calculated to be 0.4184 N and the friction force for the second block is calculated to be 0.9299 N.

The total force for the first block is 1.659 N - 0.4184 N = 1.2405 N.

The total force for the second block is 1.659 N - 0.9299 N = 0.7290 N.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the first block as 1.2405 = 0.27 x a and a = 4.59 m/s2.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the second block as 0.7290 = 0.6 x a and a = 1.215 m/s2.

4 0

3 years ago

If Zack has a mass of 80 kg calculate his weight on Earth. Please show work, thank you!

Alina [70]

784 Newton ( mass x gravity = 80 x 9.8)

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3 years ago

Which effect is created when two sound waves that are close in pitch

Temka [501]

Answer:

C. Beats

Explanation:

When waves are interfering with each other, the sound is louder in some places and softer in others. As a result, we hear pulses or beats in the sound.

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3 years ago

Emilio tries to jump to a nearby dock from a canoe that is floating in the water.Instead of landing on the dock, he falls into t

kow [346]

The reaction of him pushing off a floating object pushes it away causing him to lose power and balance

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3 years ago