A tuning fork vibrates 88 times in 0.20 s what is its frequency

Physics

1 answer:

OlgaM077 [116]3 years ago

6 0

Frequency = number of vibration/ time = 88 / 0.2 = 440 Hz

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A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

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Answer:

R=4.22*10⁴km

Explanation:

The tangential speed $v$ of the geosynchronous satellite is given by:

$v=\frac{2\pi R}{T}$

Because $2\pi R$ is the circumference length (the distance traveled) and T is the period (the interval of time).

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Since the centripetal force is the gravitational force $F_g$ between the satellite and the Earth, we know that:

$F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }$

Where G is the gravitational constant ( $G=6.67*10^{-11} Nm^{2}/kg^{2}$ ) and M is the mass of the Earth ( $M=5.97*10^{24}kg$ ). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite: