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Rina8888 [55]
3 years ago
13

Two planets are orbiting a star. Planet A is closer to the star than Planet B. Which description explains the expected motion of

Planets A and B?
Planet B orbits faster than Planet A.
Planet B has a shorter orbital period than Planet A.
Planet A orbits faster than Planet B.
Planet A has an equal orbital period to Planet B.
Physics
2 answers:
svetoff [14.1K]3 years ago
8 0

Answer: Planet A orbits faster than Planet B.

Explanation:

Kepler's laws define the motion of a planet around a star. Planets move in an elliptical orbit around the star with star at one of its foci. The speeds are governed by second law of planetary motion. The line joining planet to star sweeps equal areas in equal time intervals. This means, in order to conserve angular momentum, the planet closer to sun would orbit faster than farther planets.

It is given that planet A is closer to star than planet B. This means that planet A would orbit the star faster than B.

77julia77 [94]3 years ago
6 0
<span>Planet A orbits faster than Planet B, i believe is the answer. hope that helps

</span>
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Answer:

B

Explanation:

Given:-

- The charge of the test particle q = 3.0 * 10^-9 C

- The force exerted by the metal sphere F = 6.0 * 10^-5 N

Find:-

The magnitude and direction of the electric field strength at this location?

Solution:-

- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:  

                                       F = E*q

- Using the data given we can determine E:

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                                       E = (6.0 * 10^-5) / (3.0 * 10^-9)

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- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.

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Answer:

t = 166 years

Explanation:

In order to calculate the amount of years that electrons take to cross the complete transmission line. You first calculate the drift speed of the electrons by using the following formula:

v_d=\frac{I}{nqA}             (1)

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n: free charge density = 8.50*10^28 electrons/m^3

A: cross-sectional area of the transmission line = π*r^2

r: radius of the cross-sectional area = 2.00cm = 0.02m

You replace the values of the parameters in the equation (1):

v_d=\frac{1,010A}{(8.50*10^{28}electron/m^3)(1.6*10^{-19}C)(\pi (0.02m)^2)}\\\\v_d=5.9*10^{-5}\frac{m}{s}

Next, you use the following formula:

t=\frac{x}{v_d}                     (2)

x: length of the line transmission = 310km = 310,000m

You replace the values of vd and x in the equation (2):

t=\frac{310,000m}{5.9*10^{-5}m/s}=5.24*10^9s

Finally, you convert the obtained t to seconds

t=5.24*10^9s*\frac{1\ year}{3.156*10^7s}=166.03\ years

The electrons take approximately 166 years to travel trough the complete transmission line

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