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3 years ago

According to Kepler's third law (p2 = a3), how does a planet's mass affect its orbit around the Sun? Group of answer choices

Physics

1 answer:

anyanavicka [17]3 years ago

3 0

Answer:

A planet's mass has no effect on its orbit around the Sun.

Explanation:

The kepler's third law tells us:

$p^2=a^3$

where $p$ is the orbit period and $a$ is the semi-major axis.

As we can see from the equation, the period depends only on the measure of the semi-major axis $a$ of the orbit, that is, how far a planet is from the sun.

The equation tells us that the closer a planet is to the sun, the faster it will go around it.

The mass does not appear in the equation to calculate the period.

This is why it is concluded from the third law of Kepler that<u> the period, or the orbit of a planet around the sun, does not depend on its mass.</u>

the answer i: A planet's mass has no effect on its orbit around the Sun.

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Sound travels at a rate of 340 m/s in all directions through air. Matt rings a very loud bell at one location, and Steve hears i

Vika [28.1K]

Answer:

110 m/s

Explanation:

because if you subtract 450 from 340 you get 110

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2 years ago

A system of 4 electrons, 18 protons, and 4 neutrons has a net charge of?

slamgirl [31]

Answer:

+ 14

Explanation:

18 protons make a positive 18 charge (+18)

4 electrons make a negative 4 charge (-4)

both combined give + 18 - 4 = + 14

The four neutrons don't carry net charge, so the addition of the electrons doesn' affect the net charge found above which still gives + 14.

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3 years ago

A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a consta

dimulka [17.4K]

Answer:

0.453 m/s

Explanation:

Assuming the handle has diameter of 0.4 m while inner part diameter is 0.1 m then the circumference of outer part is $\pi d_h$ where d is diameter and subscript h denote handle. By substituting 0.4 for the handle's diameter then cirxumference of outer part is $\pi\times 0.4\approx 1.256 m$

The rate of rotation will then be 1.81/1.256=1.441 rev/s

Similarly, circumference of inner part will be $\pi d_i$ where subscript i represent inner. Substituting 0.1 for inner diameter then

$\pi\times 0.1\approx 0.3142 m$

The rate of rotation found for outer handle applies for inner hence speed will be 0.3142*1.441=0.453 m/s

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3 years ago

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$

which can also be rewritten as

$p_1 V_1 = p_2 V_2$

In our case, we have:

$p_1 = 75.9 kPa$ is the initial pressure

$V_1 = 639 cm^3$ is the initial volume

$p_2 = 125 kPa$ is the final pressure

Solving for V2, we find the final volume:

$v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3$

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4 years ago

Three identical resistors are connected in parallel to a battery. If the current of 12. A flows from the battery, how much curre

Doss [256]

Answer:

4 A

Explanation:

We are given that

$R_1=R_2=R_3=4\Omega$

I=12 A

We have to find the current flowing through each resistor.

We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.

Formula :

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$