As the moon pictured revolves around the earth, the earth also rotates. how long is a lunar cycle?

1 answer:

Volgvan3 years ago

3 0

We don't know what you mean by "lunar cycle".

It appears at the same place in your sky every 24 hours and 50 minutes on the average.

It revolved around the Earth every 27.32 days.

It displays a full set of "phases" every 29.53 days.

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Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

soldi70 [24.7K]

$Given:\\T=29.46y\approx 9.29\cdot 10^8s\\M_S\approx2.0\cdot 10^{30}kg\\G=6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2} \\\\Find:\\R=?\\\\Solution:\\\\F_g= G\frac{mM_s}{R^2} \\\\F_c= \frac{mv^2}{R} \\\\F_g=F_c\\\\G\frac{mM_s}{R^2}=\frac{mv^2}{R} \Rightarrow G\frac{M_s}{R^2}=\frac{v^2}{R}\\\\v=\omega r\\\\G\frac{M_s}{R^2}= \frac{\omega^2R^2}{R}\Rightarrow G\frac{M_s}{R^2}=\omega^2R \\\\\omega= \frac{2 \pi }{T} \\\\G\frac{M_s}{R^3}= \frac{4 \pi ^2}{T^2}$

$GM_ST^2=4 \pi ^2R^3\Rightarrow R= \sqrt[3]{ \frac{GM_ST^2}{4 \pi ^2} }\\\\\\R= \sqrt[3]{ \frac{6.67\cdot 10^{-11} \frac{m^3}{kg\cdot s^2}\cdot2.0\cdot 10^{30}kg( 9.29\cdot 10^8s)^2}{4\cdot 3.14^2} } \approx 1.42\cdot 10^{12}m$

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4 years ago

Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A p

serg [7]

Answer:

f = 735 Hz

Explanation:

given,

Person distance from speakers

r₁ = 4.1 m r₂ = 4.8 m

Path difference

d = r₂ - r₁ = 4.8 - 4.1 = 0.7 m