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LuckyWell [14K]
3 years ago
8

What measures the amount of displacement in a transverse wave

Physics
1 answer:
MAVERICK [17]3 years ago
4 0
Unlike a longitudinal wave, a transverse wave moves about, perpendicular to the direction of propagation. The particles in a transverse wave do not travel along the direction of propagation, but only oscillate up and down on its equilibrium position. With this, the displacement can be determined by measuring (in the case of electronic waves, using an oscilloscope or spectrum analyzer) and setting the desired units to measure the wave in.
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Answer:

a) F = 660.576\,N, b) a_{c} = 6.953\,\frac{m}{s^{2}}, c) v \approx 7255.423\,\frac{m}{s}, \omega = 9.583\times 10^{-4}\,\frac{rad}{s}, d) T \approx 1.821\,h

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F = G\cdot \frac{m\cdot M}{r^{2}}

F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}

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b) The centripetal acceleration of the satellite is:

a_{c} = \frac{660.576\,N}{95\,kg}

a_{c} = 6.953\,\frac{m}{s^{2}}

c) The speed of the satellite is:

v = \sqrt{a_{c}\cdot R}

v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}

v \approx 7255.423\,\frac{m}{s}

Likewise, the angular speed is:

\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}

\omega = 9.583\times 10^{-4}\,\frac{rad}{s}

d) The period of the satellite's rotation around the Earth is:

T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)

T \approx 1.821\,h

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