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3 years ago

9

Until 1883, every city and town in the United States kept its own local time. Today, travelers reset their watches only when the

time change equals 1.0 h. How far, on the average, must you travel in degrees of longitude until your watch must be reset by 9 hours? (Hint: Earth rotates 360° in about 24 h.)

1 answer:

inna [77]3 years ago

7 0

Answer:

135° (degrees)

Explanation:

The Earth rotates 360° in about 24 h

How many degrees in 1 hour

$360^{\circ}=24\ h\\\Rightarrow 1\ h=\frac{360}{24}\\\Rightarrow 1\ h=15^{\circ}$

So, in 1 hour the Earth rotates 15 degrees. This means that a person would have to reset their watch by 1 hour after travelling 15 degrees

In 9 hours

$9\ h=9\times 15^{\circ}=135^{\circ}\\\Rightarrow 9\ h=135^{\circ}$

So, a person would have to travel 135 degrees in order to reset their watch by 9 hours.

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G= constant of gravitation.

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Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

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Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

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$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

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$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

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$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

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$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

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$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

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