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Harbhajan Ibrahim Viswanathan
Asked: 2022-11-01T16:53:28+05:30 In:

Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star?

Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star?

Parabola
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  1. Data : d = 40 km, f = 1 rot/s

    ∴ r = \(\frac{d}2\) = \(\frac{40\,km}2\) = 20 km = 2 x 104 m

    Linear speed, v = ωr = (2πf)r

    = (2 × 3.142 × 1)(2 × 104) 

    = 6.284 × 2 × 104 

    = 1.257 × 10 m/s (or 125.6 km/s)

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Radheshyam Panchal
Asked: 2022-10-31T09:10:39+05:30 In:

Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star?

Certain stars are believed to be rotating at about 1 rot/s. If such a star has a diameter of 40 km, what is the linear speed of a point on the equator of the star?

Parabola
Leave an answer

Leave an answer

Browse

1 Answer

  1. Data : d = 40 km, /= 1 rot/s

    ∴ r = \(\frac{d}2\) = \(\frac{40\,km}2\) = 20 km = 2 × 104 m

    Linear speed, v = ωr = (2πf)r 

    = (2 × 3.142 × 1)(2 × 104) 

    = 6.284 × 2 × 104

    = 1.257 × 105 m/s (or 125.6 km/s)

    • 0