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Hitman42 [59]
3 years ago
8

Physics homework help.

Physics
1 answer:
Monica [59]3 years ago
8 0
(24.6 m/s) / (50 m/s^2)

= 0.492 second.
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Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1
hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

So:

\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0
3 years ago
Importance of electrical symbol​
Vsevolod [243]
The electrical symbols are very important especially when fixing electrical appliances because it tells you where the wire of neutral and live go.
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4 years ago
n oscillator is driven by a sinusoidal force. The frequency of the applied force A : must be less than the natural frequency of
lilavasa [31]

Answer:

  B : is independent of the natural frequency of the oscillator

Explanation:

You can apply any force you like to a natural oscillator. It is independent of the natural frequency of the oscillator.

The result you get will depend on how the frequency of the applied force and the natural frequency relate to each other. It will also depend on the robustness of the oscillator with respect to the applied force.

Clearly, if the force is small enough, it will have no effect on the oscillator. If it is large enough, it will overpower any motion the oscillator may attempt. For forces in the intermediate range, there will be some mix of natural oscillation and forced behavior. One may modulate the other, for example.

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3 years ago
What is the answer to this definition ?
Paul [167]

Newton's 2nd law of motion is:    <em>Force</em> = (mass) x (acceleration) .

Force is the only way to change an object's velocity.


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Which statement is true of a concave lens?
natka813 [3]
It produces only virtual images is the answer
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