Answered

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What is the approximate tangential speed of an object orbiting earth with a radius of 1. 8 × 108 m and a period of 2. 2 × 104 s? 7. 7 × 10–4 m/s 5. 1 × 104 m/s 7. 7 × 104 m/s 5. 1 × 105 m/s.

Sagot :

Answer: the period is 2.2 x 104 s, then the rotational speed is:

1 / 2.2x104 = 4.55x10-5 rad/s

We can solve for the tangential speed using the formula:

v = ωr

where ω is the rotational speed

v is the linear or tangential speed

r is the radius

So,

v = 4.55x10-5(1.8x108)

v = 5.1 x 104 m/s

Explanation:

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