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melisa1 [442]
3 years ago
14

Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's

laws to predict the orbital period of such a planet.
Physics
1 answer:
mamaluj [8]3 years ago
4 0

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

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A turntable that spins at a constant 80.0 rpmrpm takes 3.50 ss to reach this angular speed after it is turned on. Find its angul
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Answer:

The angular acceleration is <u>2.39 rad/s²</u>.

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Time taken (t) = 3.50 s

First, let us determine the final angular speed in radians per second.

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Now, using equation of motion for rotational motion, we have:

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Plug in the given values and solve for α. This gives,

8.38=0+\alpha \times 3.50\\\\\alpha=\frac{8.38}{3.50}=2.39\ rad/s^2

Therefore, the angular acceleration is 2.39 rad/s².

Now, again using rotational equation of motion relating angular displacement, we have:

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Plug in the given values and solve for 'θ'. This gives,

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Answer:

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When changing from liquid to gas (for example when water evaporates), the heat necessary for this to happen is called latent heat of vaporization. The word latent means hidden, because a change in temperature is not perceived during the phase change, even when heat is being added, thus it is said that the heat is hidden or latent.

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When three people with a total mass of 2.00 x 102 kg step into their 1.200 x 103 kg car, the car’s
Aleksandr-060686 [28]

Answer:

a

 k =  457333.3 N/m

b

x_a  =0.09\ m      

Explanation:

From the question we are told that

    The total mass of  three people is  M  = 2.00*10^{2} \ kg

     The mass of the car is  m_c  =  1.200 *10^{3} \ kg

     The compression of the car spring is  x = 3 \ cm  = 0.03 \ m

     

Generally the spring constant is mathematically represented as

          k =  \frac{F}{x}

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=>       F = (M +m_c) *g

=>       F = ([2.0*10^{2} ]+[ 1.200*10^{3}]) * 9.8

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        k =  \frac{13720}{0.03}

=>     k =  457333.3 N/m

Generally if the mass which the car is loaded with is  m  =  3.00*10^{2} \ kg

Then the force experienced by the spring is  

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         =>       F_a = (3.00*10^{3} + 1.200 *10^{3}) * 9.8

         =>       F_a = 41160 \  N

Generally from the above formula the compression is  

       x_a  = \frac{F_a}{k}

=>    x_a  = \frac{41160}{457333.3}

=>    x_a  =0.09\ m      

3 0
3 years ago
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