The satellite s travels around the earth in a circular path with a constant speed of 20 mm>h. If the acceleration is 2.5 m&gt

;s2, determine the altitude h. Assume the earth's diameter to be 12 713 km.

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kicyunya [3294]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = $60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s$

s = Distance traveled = $123\ ft=\dfrac{123}{3.281}=37.4885705578\ m$

$\theta$ = Incline = $26^{\circ}$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2$

Friction coefficient

$\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561$

$mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m$

The calculated stopping distance is 83.1946504051 m

6 0

4 months ago

serg [3582]

Answer:

Insufficient details provided; please clarify further.

5 0

4 months ago