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A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.90s apart. The speed of sound in air is343 m/s, and in concrete is 3000 m/s.How far away did the impact occur?

Respuesta :

The impact is occurred at a distance 348.55 m far from person.

Explanation:

Let s be the distance to the impact position and t be the time when he hears the sound though concrete after impact.

Time when he hears impact through air = t + 0.90

The distance traveled by sound wave in both concrete and air is s.

Speed of sound in air = 343 m/s

Speed of sound in concrete = 3000 m/s

Distance traveled in air = Distance traveled in concrete

343 x Time when he hears impact through air = 3000 x Time when he hears impact through concrete

        343 x (t + 0.90) = 3000 x t

                  3000t - 343 t = 343 x 0.90

                  2657t = 308.7

                      t = 0.116 s

Distance traveled in concrete = 3000 x t = 3000 x 0.116 = 348.55m

So the impact is occurred at a distance 348.55 m far from person.

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