Question

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 0.70 ss apart. The speed of sound in air is 343 m/sm/s, and in concrete is 3000 m/sm/s. Part A How far away did the impact occur

Answer 1

271.095 m

Step-by-step explanation:

✓ Let speed of sound in air that was given as (343 m/s) be represented as (Vi)

✓( speed of sound in concrete that was given as (3000 m/s ) be debited as (Vc)

✓ Let the distance travelled by the sound = s

✓duration of Time that exist between heard of sounds = 0.70s

But we know that

Time = (Distance / Speed)

✓Time it takes the sound to travel through air= s/vi = s/343

✓Time it takes the sound to travel through concrete= s/vc = s/3000

✓ (s/343) - (s/3000) = 0.70

Finding LCM and simplify

[(3000s - 343s)]/1029000 = 0.70

2657s /1029000 = 0.70

Making " s" subject of the formula

s= (1029000 × 0.70)/2657

s=720300/ 2657

s= 271.095 m

Hence, The impact took place at a distance of 271.095 m away from the person.