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3 years ago

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Physics

1 answer:

vekshin13 years ago

4 0

Answer:

At a convergent boundary where both plates are continental, mountain ranges grow and earthquakes are common. At a transform boundary, there is a transform fault and massive earthquakes occur but there are no volcanoes. Processes acting over long periods of time create Earth's geographic features.

Explanation:

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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

The allowed energies of a simple atom are 0.0 eV, 4.0 eV, and 6.0 eV. Part A What wavelength(s) appear(s) in the atom's emission

Dmitriy789 [7]

Answer:

3.1 × 10^- 7 m and 2.1 × 10^-7 m

Explanation:

First we must convert each value of energy to Joules by multiplying its value by 1.6 ×10^-19. After that, we can now obtain the wavelength from E= hc/λ

Where;

h= planks constant

c= speed of light

λ= wavelength of light

For 6.0ev;

E= 6.0 × 1.6 ×10^-19

E= 9.6 × 10^-19 J

From

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/9.6 × 10^-19

λ= 2.1 × 10^-7 m

For 4.0 eV

4.0 × 1.6 × 10^-19 = 6.4 × 10^-19 J

E= hc/λ

λ= hc/E

λ= 6.6 × 10^-34 × 3 × 10^8/6.4 × 10^-19

λ= 3.1 × 10^- 7 m

3 0

3 years ago

The force of gravity is an inverse square law. this means that, if you double the distance between two large masses, the gravita

likoan [24]

F~1/r²
doubling the distance r, Decreases the force by ¼

4 0

3 years ago

A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along

storchak [24]

Answer:

$v = \frac{kQ}{a}$

Explanation:

We define the linear density of charge as:

$\lambda = \frac{Q}{L}$

Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

$dv = \frac{kdq}{r}$

where k is the coulomb's constant

r is the distance from dq to center of the circle

Thus.

$v = \int_{}^{}\frac{kdq}{a}$

$v = \frac{k}{a}\int_{}^{}dq$

$v = \frac{kQ}{a}$ Potential at the center of the semicircle

4 0

3 years ago

1. A Faraday ice pail similar to the one you will use in the experiment is used in this question. The principle of its operation

algol [13]

Answer:

your mommy

Explanation:

YOUR MAMMA

I LOVE MY DADDY

HE DOES ME REAL HARD

8 0

3 years ago