A certain organ pipe, open at both ends, produces a fundamental frequency of 300 Hz in air. If the pipe is filled with helium at
1 answer:
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(a) -83.6 Hz
Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:
where
f' is the apparent frequency
v = 343 m/s is the speed of sound
is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)
f is the original frequency of the sound
Here we have
f = 350 Hz
When the train is approaching, we have
So the frequency heard by the observer on the platform is
When the train has passed the platform, we have
So the frequency heard by the observer on the platform is
Therefore the overall shift in frequency is
And the negative sign means the frequency has decreased.
(b) 0.865 m
The wavelength and the frequency of a wave are related by the equation
where
v is the speed of the wave
is the wavelength
f is the frequency
When the train is approaching the platform, we have
v = 343 m/s (speed of sound)
f = f' = 396.7 Hz (apparent frequency)
Therefore the wavelength detected by a person on the platform is
Answer:
a. The ratio of their resistance is 2783:64
b. The ratio of their energy is 4:23
c. The charge on the first bulb is 5.75 C
The charge on the second bulb is C
Explanation:
The voltage on one of the electric bulbs, V₁ = 40 volts
The power rating of the bulb, P₁ = 230 w
The voltage on the other electric bulbs, V₂ = 110 volts
The power rating of the bulb, P₂ = 40 w
a. The power is given by the formula, P = I·V = V²/R
Therefore, R = V²/P
For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96
The resistance of the second bulb, R₂ = 110²/40
The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64
∴ The ratio of their resistance, R₂:R₁ = 2783:64
b. The energy of a bulb, E = t × P
Where;
t = The time in which the bulb is powered on
∴ The energy of the first bulb, E₁ = 230 w × t
The energy of the second bulb, E₂ = 40 w × t
The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23
∴ The ratio of their energy, E₂:E₁ = 4:23
c. The charge on a bulb, 'Q', is given by the formula, Q = I × t
Where;
I = The current flowing through the bulb
From P = I·V, we get;
I = P/V
For the first bulb, the current, I = 230 w/40 V = 5.75 amperes
The charge on the first bulb per second (t = 1) is therefore;
Q₁ = 5.75 A × 1 s = 5.75 C
The charge on the first bulb, Q₁ = 5.75 C
Similarly, the charge on the second bulb, Q₂ = (40 W/110 V) × 1 s = C
The charge on the second bulb, Q₂ = C.
d. The question has left out parts