Answer:
You will hear the note E₆
Explanation:
We know that:
Your speed = 88m/s
Original frequency = 1,046 Hz
Sound speed = 340 m/s
The Doppler effect says that:
Where:
f = original frequency
f' = new frequency
v = velocity of the sound wave
v0 = your velocity
vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.
Replacing the values that we know in the equation we have:
This frequency is close to the note E₆ (1,318.5 Hz)
Answer:
The distance of the goggle from the edge is 5.30 m
Explanation:
Given:
The depth of pool (d) = 3.2 m
let 'i' be the angle of incidence
thus,
i =
i = 67.75°
Now, Using snell's law, we have,
n₁ × sin(i) = n₂ × 2 × sin(r)
where,
r is the angle of refraction
n₁ is the refractive index of medium 1 = 1 for air
n₂ is the refractive index of medium 1 = 1.33 for water
now,
1 × sin 67.75° = 1.33 × sin(r)
or
r = 44.09°
Now,
the distance of googles = 2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) = 5.30 m
Hence, <u>the distance of the goggle from the edge is 5.30 m</u>