Answer:
1) The speed of sound increases
2) 440 Hz
3) 29°C
4) 17°C
5) 434 Hz
6) 12 m/s
7) 17.3 m
Explanation:
1) The speed of sound increases
2) V = f×λ
f = V/λ = 343/0.78 = 439.744 ≈ 440 Hz
3) V = f×λ
512 × 0.68 = 348.16 m/s
348.16 - 331 = 17.16
T = 17.16/0.6 = 28.6 ≈ 29°C
4) Increase in speed = 350 - 340 = 10
Increase in temperature = 10/0.6 = 16.67° ≈ 17°C
5) f = V/λ = 343/0.79 = 434 Hz
6) 331 + 0.6×30 - (331 × 0.6 ×10) = 12 m/s
7) V = 331 + 0.6×25 = 346m/s
λ = 346/20 = 17.3 m
Y, bc the height of the bounce back is higher than x
Answer:
3.2m
Explanation:
Given parameters:
Frequency of the FM radio = 9.23 x 10⁷Hz
Velocity of the waves = 3 x 10⁸m/s
Unknown:
Wavelength of the wave = ?
Solution:
To solve for the wavelength of the wave, we need the velocity equation;
Velocity = frequency x wavelength.
Radio waves are all electromagnetic radiations produced by both electrical and magnetic fields perpendicularly oriented to one another.
Since the unknown is wavelength, we solve for it:
3 x 10⁸ = 9.23 x 10⁷ x wavelength
wavelength = 
wavelength = 3.2m
The Doppler effect is the right concept to solve this problem. The Doppler effect is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be described as,

Here,
= Frequency of the sound from the Whistle
f = Frequency of sound heard
v = Speed of the sound in the Air
Replacing we have that





Therefore the minimum speed to know if the whistle is working is 16.33m/s