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Brilliant_brown [7]
3 years ago
12

Twoâ cars, 405 kilometersâ apart, begin driving toward each other on aâ long, straight highway. one car travels 70 kilometers pe

r hour and the other 65 kilometers per hour. at the sameâ time, aâ canary, starting on oneâ car, flies back and forth between the two cars as they approach each other. if the canary flies 100 kilometers per hour and spends no time to turn around at eachâ car, how far has it flown when the carsâ collide?
Physics
1 answer:
mart [117]3 years ago
8 0
The space between the cars closes at (70+65)=135 km/hr. They collide in (405/135)=3 hours after they start moving. In 3 hours, the canary covers (100km/hr x 3)= 300 km.
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If an object has a mass of 50 kg and is accelerating at 2m/s/s, what is the force it will exert?
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The formula you're looking for is Force = mass * acceleration. this means the force = 50 * 2, which is 100N
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Two identical stars with mass M orbit around their center of mass. Each orbit is circular and has radius
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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.

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