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Dmitrij [34]
2 years ago
9

Suppose astronomers discover a new planet farther away from the Sun than Earth. How would the day and year of this planet compar

e to Earth's?
A. The planet's day would be longer, and its year would be shorter.
B. The planet's day would be shorter, and its year would be longer.
C. The planet's year would be longer, but it is impossible to predict how its day would compare to Earth's.
D. The planet's year would be shorter, but it is impossible to predict how its day would compare to Earth's. PLEASE HELP ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️
Physics
2 answers:
Volgvan2 years ago
8 0
The year would definitely be longer, due to Keller’s 3rd law, but I know of several objects orbiting farther from the sun than earth with both much longer and much shorter day lengths.
posledela2 years ago
6 0
I think the answer would be C. because the orbit around the sun compared to the Earth's (Identifies the seasons change and the year) would be longer, but the day is unknown because that is where it would come down to the total mass of the planet and other important factors about the planets properties.
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Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous
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Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

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Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

A_1V_1=A_2V_2

A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}

the diameter decreases 86% so

d_2 = 0.86d_1

v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}

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Answer:

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Explanation:

It is given that,

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Let v is the resultant of velocity. It is given by :

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Let E is the kinetic energy of the plane. It is given by :

E=\dfrac{1}{2}mv^2

E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2

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Answer:

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