Question

Question 7 (1 point) While returning to school after an extended summer break, you and a physically distant friend wonder how far you are from each other. You decide to use your summer school physics lesson. Your smart watch reads 23°C, and you manage to record that a clap in your friends direction is heard 8.0 x 103 s later. Are you and your friend the required 2m apart? [WWNBHU]

Answer 1

Given that temperature at the location is 23 C

Now speed of sound at T temperature is given by:

V = 331 + 0.6*T

Since T = 23 C, So

V = 331 + 0.6*23 = 344.8 m/s

Now given that clap is heard after 8.0*10^-3 sec, So

We know that:

time = distance between both persons/speed of sound

d = V*t

d = 344.8*8.0*10^-3

d = 2.7584 m = distance between both friends

So no you and your friend are more than 2 m apart, where actual distance between you and your friend is 2.7584 m

Let me know if you've any query.