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melomori [17]
3 years ago
5

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

, in astronomical units, AU. If planet Y is twice the mean distance from the sun as planet X, by what factor is the orbital period increased? A. 2^1/3. B. 2^1/2. C. 2^2/3. D. 2^3/2
Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

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Answer:

P = 133.13 Watt

Explanation:

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final angular speed after friction is given as

\omega_f = 2\pi f

\omega_f = 2\pi(7.5/3600)

\omega_f = 0.013 rad/s

now angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{0.015 - 0.013}{15}

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now torque due to friction on the wheel is given as

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P = \tau \omega

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8 0
3 years ago
An inventor claims to have developed a power cycle having a thermal efficiency of 40%, while operating between hot and cold rese
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From Carnot's theorem, for any engine working between these two temperatures:


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7 0
3 years ago
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The time for the car to drive directly south is determined as 7.15 s.

<h3>Time for the car to drive directly west</h3>

The time for the car to drive directly south is determined by applying the concept of slope.

slope = Δy/Δx

a = Δv/Δt

Δt = Δv/a

Δt = (26.8)/(3.75)

Δt = 7.15 s

Thus, the time for the car to drive directly south is determined as 7.15 s.

Learn more about acceleration here: brainly.com/question/605631

#SPJ1

7 0
2 years ago
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Answer:

Explanation:

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