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Suppose an eruptive prominence rises at a speed of 150 km/s. If it does not change speed, how far from the photosphere will it extend after 3 hours? How does this distance compare with the diameter of Earth?

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Answer:

127 times

Step-by-step explanation:

v = speed of eruptive prominence = 150 km/s = 150,000 m/s

t = time interval = 3 hours = 3 x 60 x 60 sec = 10800 sec

d = distance traveled by photosphere

Distance traveled is given as

d = v t

d = (150,000) (10800)

d = 1.62 x 10⁹ m

d = 1.62 x 10⁶ km

D = diameter of earth = 12742 km

n = number of times the distance traveled by eruptive prominence

using the equation

d = n D

1.62 x 10⁶ = n (12742)

n = 127 times

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