On a bright winter's day, a student notices an echo as she runs between a high school's playing field and the rear wall of its gym. she decides to estimate how far she stands from the reflective wall. she abruptly whistles and determines that the sound pulse returns to her as an echo 0.431 seconds later (averaging several trials for enhanced reliability). the air is a frigid -10°c. six months later, the student decides to repeat her experiment on a hot summer day, standing at the same exact spot. this time, the echo takes a slightly shorter 0.396 seconds to return to the student. (a) what is the temperature (in °c) of the air on the summer day? oc (b) how far (in m) does the student stand from the reflective walir?

(a)

(b)

Explanation:

Speed of Sound

The sound travels in an open space at a predictable speed which mainly depends on the air temperature. It can be calculated with an approximate formula in terms of the temperature T in degrees Celsius.

The speed comes in m/s

If we have determined the speed of sound, it's possible to estimate the distance to a large obstacle that could cause a returning sound (echo). The formula for the distance is

Where is the speed of sound and t is half the time we hear our echo, because we can measure the forth and back times, assumed to be equal.

(b) We know the air is at -10^oC, so the speed of sound is

The echo was heard 0.431 seconds later, so t=0.431 / 2= 0.2155 seconds

The distance to the rear wall of the gym is

(b) On a summer day, the experiment was repeated and the echo was heard at t = 0.396 seconds. Since we already know the distance to the wall, we can estimate the temperature of the air, by solving the equation for T

We need to compute the new value of , with

We know

Now we can find T

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west