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anyanavicka [17]

3 years ago

12

An airplane travels directly from Washington, D.C., to Atlanta, Georgia, a distance of 850 km at a velocity of 425 km/h southwes

t
How long does the trip take in hours?

1.0 h

2.0 h

0.5 h

4.0 h

1 answer:

Sloan [31]3 years ago

4 0

Answer:

obviously 2 hours cuz 850/425

Explanation:

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A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force e

Natali5045456 [20]

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) $T \approx 1.821\,h$

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

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

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

$F = 660.576\,N$

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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