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beks73 [17]
3 years ago
13

PLZ HELP ME!!!!!!!!! WILL GIVE BRAINLY!!!!!!!! Which photo represents a waxing gibbous moon?

Physics
1 answer:
yarga [219]3 years ago
6 0
A phase of moon with more than half sunlit portion visible on the right
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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.

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3 years ago
The Doppler effect occurs when a source of sound moves
Dmitrij [34]

The Doppler effect occurs when a source of sound or light
moves either toward or away from the observer.

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3 years ago
A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the 14) car a deceleration of 3.50
sergij07 [2.7K]
To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving. 

First, we manipulate the one of the kinematic equations
 
v^2 = v0^2 + 2 (a) (x)  where v = 0 since the car stopped

Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,

<span>we get v0 = sqrt (2(a)(x))

Substituting the known values,

v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s 
</span>
Therefore, before stopping the car the original speed of the car would be 14.49 m/s
7 0
4 years ago
in the circuit diagram even in suppose the resistor R1 R2 and R3 have the value of 5 ohm 10 ohm 30 ohm respectively which have b
Anuta_ua [19.1K]
<h3><u>Given</u> :</h3>

Three identical resistors of resistances 5Ω, 10Ω and 30Ω are connected with a battery of 12V.

<h3><u>To Find</u> :</h3>

We have to find current through the each resistor and equivalent resistance of circuit.

<h3><u>SoluTion</u> :</h3>

➝ Equivalent resistance of series connection is given by

  • <u>R = R1 + R2 + R3</u>

➝ We know that, Equal current flow through each resistor in series connection.

➝ As per ohm's law, Current flow through a conductor is directly proportional to the applied potential difference.

  • <u>V = IR</u>

◈ <u>Equivalent resistance</u> :

⇒ Req = R1 + R2 + R3

⇒ Req = 5 + 10 + 30

⇒ <u>Req = 45Ω</u>

◈ <u>Current flow in circuit</u> :

⇒ V = IReq

⇒ 12 = I × 45

⇒ <u>I = 0.27A</u>

፨ Therefore, 0.27A current will flow through each resistor.

5 0
3 years ago
Acceleration is a derived unit ​
aleksklad [387]

Acceleration is a derived unit because it is derived from two quantities : Velocity and time.

We know, acceleration = Change in velocity/Time

5 0
2 years ago
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