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3 years ago

Kepler's third law can be used to compare the moon's orbit around the earth to the orbit of the earth around the sun.

Physics

1 answer:

Pavlova-9 [17]3 years ago

5 0

No, you can't! The planet should rotate around a common body!

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Consider two planets of mass m and 2m,

Rzqust [24]

Answer:

Part a)

$\frac{F_1}{F_2} = 10.125$

Part b)

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

$\frac{T_1}{T_2} = 9.54$

Explanation:

Part a)

As we know that the gravitational force is given as

$F = \frac{GMm}{r^2}$

so we will have to find the ratio of force on two planets due to star

so here we have

$\frac{F_1}{F_2} = \frac{m_1r_2^2}{m_2r_1^2}$

$\frac{F_1}{F_2} = \frac{m (4.5r)^2}{(2m) r}$

$\frac{F_1}{F_2} = 10.125$

Part b)

Orbital speed is given as

$v = \sqrt{\frac{GM}{r}}$

so the ratio of two orbital speed is given as

$\frac{v_1}{v_2} = \frac{r_2}{r_1}$

$\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12$

Part c)

Time period is given as

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

so the ratio of two time period is given as

$\frac{T_1}{T_2} = \sqrt{\frac{r_1^3}{r_2^3}}$

$\frac{T_1}{T_2} = \sqrt{\frac{4.5r^3}{r^3}}$

$\frac{T_1}{T_2} = 9.54$

8 0

3 years ago

Read 2 more answers

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

4 years ago

A motarcycle covers a distance of 1.8km in 5minutes.Calculate its average velocity

MissTica

Answer:

0.6 m/s

Explanation:

1.8 km = 1800 m

5 minutes= 300 s

Now,

Velocity = Displacement / time

= 1800/300

= 0.6 m /s

8 0

3 years ago

A wye-connected load has a voltage of 480v applied to it. What is the voltage drop across each phase

rodikova [14]

Answer:

$Y_A=277.128 \angle 30v$

$Y_B=277.128 \angle (-150)v$

$Y_C=277.128 \angle (90)v$

Explanation:

From the question we are told that

Voltage $V_L_L =480v$

Generally in a case of Y_connection $V_p_ h$ is mathematical represented as

$V_p_h=\frac{V_l_l}{\sqrt{3}} \angle (\phi-30)v$

Generally voltage drop across phase A

$Y_A=\frac{408}{\sqrt{3}} \angle -(0-30)$

$Y_A=277.128 \angle 30v$

Generally voltage drop across phase B

$Y_B=277.128 \angle (-30-120)$

$Y_B=277.128 \angle (-150)v$

Generally voltage drop across phase C

$Y_C=277.128 \angle (-30+120)$

$Y_C=277.128 \angle (90)v$

3 0

3 years ago

What is the speed of a wave if the wavelength is 3 meters and the frequency is 200hz

RideAnS [48]

Wave speed = frequency * wavelength

Input the numbers into this equation :

Wave speed = 200 * 3

Work it out and you will get the answer :

Wave speed = 600 m/s

6 0

3 years ago