All categories

2 years ago

An asteroid has a mean radius equal to 17 times that of Earth. Using Kepler's 3rd law find its period. T^2=a^3

Physics

1 answer:

vagabundo [1.1K]2 years ago

8 0

Answer:

If T^2 = a^3

T1^2 / T2^2 = a1^3 / a2^3

Or T2^2 = T1^2 * ( a2^3 / a1^3)

Given T1 = 1 yr and a2 / a1 = 17

Then T2^2 = 1 * 17^3

or T2 = 70 yrs

You might be interested in

A rigid container holds 0.30g of hydrogen gas.

Strike441 [17]

Answer:

Part A: $\mathbf{Q =94 \ J}$ to two significant figures

Part B: $\mathbf{Q =160 \ J}$ to two significant figures

Part C: $\mathbf{Q =220 \ J}$ to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = $C_v=\dfrac{3}{2}R$

For Part A:

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K$

$Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)$

$Q=93.5325 \ J$

$\mathbf{Q =94 \ J}$ to two significant figures

Part B. For hot temperature, $C_v=\dfrac{5}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K$

$Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)$

$Q=155.8875 \ J$

$\mathbf{Q =160 \ J}$ to two significant figures

Part C. For an extremely hot temperature, $C_v=\dfrac{7}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K$

$Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)$

$Q=218.2425 \ J$

$\mathbf{Q =220 \ J}$ to two significant figures

6 0

2 years ago

Samantha is viewing a real image formed by a lens of a diamond ring. Which must be true about the image?

Ad libitum [116K]

<span>It could not be captured on film. is the answer</span>

4 0

2 years ago

Read 2 more answers

A bare 4 AWG copper conductor Installed horizontally near the bottom or vertically, and within that portion of a concrete foun d

neonofarm [45]

Answer:

20 ft

Explanation:

This is the length where the direct contact could be used as ground electrode

4 0

3 years ago

A boxer can hit a heavy bag with great force. Why can't he hit a piece of tissue paper in midair with the same amount of force?

12345 [234]

Answer:

This is due to impulse

Explanation:

Impulse equal to mΔv and FΔt

You can set these equal as mΔv = FΔt

When a boxer punches a tissue, it is like punching a cushion or a pillow. The time that the hit takes is much grater than if they were to hit something solid. In addition, the change in velocity of the boxer's arm would be much greater when they hit a punching bag. In this equation, the greater the time, the less force that is needed.

6 0

2 years ago

According to Newton's law of universal gravitation, which statements are true?

kogti [31]

Answer:

c. As we gain mass, the force of gravity on us increases

8 0

2 years ago

Read 2 more answers