On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth’s atmosphere over Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. The blast wave traveled at 10° above the horizon. (a) What was the average velocity of the blast wave? b) Compare this with the speed of sound, which is 343 m/s at sea level.

Question

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Respuesta :

lucic lucic · Answer 1 · 2019-10-06T20:15:59+02:00

Answer:

a)156.7 m/s

b)the speed of sound is twice that of the superbolide.

Explanation:

Given in the question;

Explode attitude of superbolide=23.5 km

The time taken by blast wave is 2.5 minutes

Blast wave traveling angle is 10° above the horizon

a).

Average velocity of blast is given by;

Δx/Δt where Δx is change distance traveled and Δt is time taken

Δx=23.5 × 1000 = 23500 m

Δt= (2×60)+30 = 120+30 =150 seconds

v=23500/150 = 156.7 m/s

b) Comparing the speeds

Speed of sound is 343m/s

Speed of superbolide= 156.7m/s

Finding the ratio of speed of sound to that of the superbolide

=343/156.7 =2.18

Hence the speed of sound is twice that of the superbolide.