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umka21 [38]
3 years ago
5

n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011

m. How many earth days does it take for the planet ot execute one complete orbit about the sun
Physics
1 answer:
Anna35 [415]3 years ago
4 0

Answer:

F = M2 ω^2 R       centripetal force of sun on planet

ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P        where P is the period

P = 2 pi (M2 * R / F)^1/2

F = G M1 M2 / R^2        gravitational force on planet

P = 2 pi {R^3 / (G M1)]^1/2

P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2

P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec

1 yr = 3600 * 24 * 365 = 3.15E7 sec

P = 3.9 / 3.2 = 1.2 years

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4 years ago
A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.
Norma-Jean [14]

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2.1844 m/s

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The principle of conservation of momentum can be applied here.

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0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\

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0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}

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Answer:

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