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RoseWind [281]
3 years ago
15

How fast must a plane fly along the earth's equator so that the sun stands still relative to the passengers? Give your answer in

both km/h and mph.The radius of the earth is 6400 km.
part A km/h
part B mph
Part C In which direction must the plane fly, east to west or west to east?
Express your answer using two significant figures.
Physics
1 answer:
Mamont248 [21]3 years ago
4 0

Answer:

17×10² km/h

10×10² mph

east to west

Explanation:

Circumference of Earth

2\pi r\\ =2\pi 6400000\\ =12800000\pi\ m=12800\pi km

Speed = Distance / Time

Speed=\frac{12800\pi}{24}=1675.51\ km/h

Converting to mph

1\ km=\frac{1}{1.60934}\ mi

1675.51\ km=1675.51\times \frac{1}{1.60934}\ mi=1041.11\ mi

The velocity of the plane must be 17×10² km/h or 10×10² mph

The plane must fly east to west

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Answer:

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Explanation:

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angular velocity and period are related

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we substitute

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In this exercise indicate that for a long Lo the period is To, then and increase the long

          L = L₀ + ΔL

we substitute

           T = 2\pi  \ \sqrt{\frac{L + \Delta L}{g} }

            T = 2\pi  \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }

in general the length increments are small ΔL/L «1, let's use a series expansion

           \sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...  

we keep the linear term, let's substitute

           T = 2\pi  \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L}  )  

if we do

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           T₀ + ΔT = 2\pi  \sqrt{\frac{\Delta L}{g} }  + \pi  \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}

            T₀ + ΔT = T₀ + \pi  \sqrt{\frac{1}{Lg} } \ \Delta L

            ΔT = \pi \ \frac{\Delta L}{\sqrt{Lg} }

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