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stellarik [79]
3 years ago
5

Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a st

opwatch just as the top of the sun disappears. You then stand, elevating your eyes by a height H = 1.43 m, and stop the watch when the top of the sun again disappears. If the elapsed time is it = 11.9 s, what is the radius r of the planet to two significant figures? Notice that duration of a solar day at the far-off planet is the same that is on Earth.

Physics
1 answer:
zubka84 [21]3 years ago
6 0

Answer:

R=3818Km

Explanation:

Take a look at the picture. Point A is when you start the stopwatch. Then you stand, the planet rotates an angle α and you are standing at point B.

Since you travel 2π radians in 24H, the angle can be calculated as:

\alpha =\frac{2*\pi *t}{24H}  t being expressed in hours.

\alpha =\frac{2*\pi *11.9s*1H/3600s}{24H}=0.000865rad

From the triangle formed by A,B and the center of the planet, we know that:

cos(\alpha )=\frac{r}{r+H}  Solving for r, we get:

r=\frac{H*cos(\alpha) }{1-cos(\alpha) } =3818Km

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0.05312 m

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V_f,p = - (m_b /m_p) * V_f,b

V_f,p = - (5.76 / 184)*V_f,b

V_f,p = - 0.0313*V_f,b   ....1

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d = -0.05312 m

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Please help! If you held a piece of tissue paper at arms length and asked the heavyweight champion of the world to hit the paper
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