The answer is A. Hope this helps!!!
Sound level at distance of 15 m is given as 20 dB
so intensity at this distance is given as
now if we move closer to some some distance the sound level is now 50 dB
now the intensity is given as
now we know that
so now the distance from friend must be 47 cm
Answer: f = 927.55Hz
Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below
L = λ/4 where λ = wavelength.
speed of sound in air = v = 343m/s.
fundamental frequency of open closed tube = 315Hz
λ = 4L.
v = fλ
343 = 315 * 4L
343 = 1260 * L
L = 343/ 1260
L = 0.27m
In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.
The length of tube and wavelength are related by the formulae below
L = λ/4, λ=4L
λ = 4 * 0.27
λ = 1.087m.
v = fλ
1010 = f * 1.087
f = 1010/1.807
f = 927.55Hz
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
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