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3 years ago

Plzz answer this question correctly

Physics

1 answer:

Lelechka [254]3 years ago

6 0

Answer:

I'm not sure sorry.... shshsjsk

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An object moves in simple harmonic motion described by the equation d equals one fifth sine 2 t where t is measured in seconds a

tensa zangetsu [6.8K]

Answer:

The maximum displacement, the frequency, and the time required for one cycle are 1/5 inch, 1/π Hz and π sec.

Explanation:

Given that,

The equation of simple harmonic motion

$d = \dfrac{1}{5}\sin 2t$ .....(I)

We need to calculate the maximum amplitude

Using equation of simple harmonic motion

$y = a \sin\omega t$

Where, a = amplitude

$\omega$ =frequency

t = time

On comparing equation (I) and general equation

The amplitude is a maximum displacement traveled by a wave.

$a = \dfrac{1}{5}$

So, the maximum displacement is

$d= \dfrac{1}{5}\ inch$

We need to calculate the frequency

$\omega=2\pi f$

$f = \dfrac{1}{\pi}\ Hz$

We need to calculate the time required for one cycle

$t =\dfrac{1}{f}$

$t =\pi\ sec$

Hence, The maximum displacement, the frequency, and the time required for one cycle are 1/5 inch, 1/π\ Hz and π\ sec.

3 0

3 years ago

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.