An alarm clock is dropped off the edge of a tall building. You, standing directly under it, hear a tone of 1350 Hz coming from the clock at the instant it hits the ground. Since you know the building is 25 m tall, you can find out what the frequency of the alarm would be if you had just held it in your hands. What would that frequency be

the frequency clock would be 1262.85 Hz

Explanation:

Given data;

height of building h = 25 m

from the third equation of motion;

v² = u² + 2as

Since the Alarm clock falls with an acceleration equal to the acceleration due to gravity; a = g = 9.81 m/s²

initial velocity u = 0

so we substitute our values into the kinematic equation

v² = (0)² + 2 × 9.81 × 25

v² = 490.5

v = √490.5

v = 22.1472 m/s

Now, since the alarm clock is moving both I am stationary;

my velocity will be zero.

so Frequency of the alarm clock will be;

f' = [ (v - ) / ( v + ) ] × f

we know that; speed of sound is 343 m/s, so v = 343 m/s, is 22.1472 m/s, f is 1350 Hz,  is 0 m/s

so we substitute the values into the equation

f' = [ (343 - 22.142 ) / ( 343 + 0 ) ] × 1350

f' = [ 320.858 / 343 ] × 1350

f' = 0.935446 × 1350

f' = 1262.85 Hz

Therefore, the frequency clock would be 1262.85 Hz

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.