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-Dominant- [34]
3 years ago
6

A pipe that is open at both ends and a tuning fork of frequency 240 Hz are being used as a fancy new thermometer, albeit one tha

t utilizes sound interference. At room temperature (20.0°C), the pipe's fundamental frequency and the tuning fork's frequency are identical and thus no beat frequency is heard between the two. When the temperature rises to 43.0°C, what beat frequency (in Hz) should be heard?
Physics
1 answer:
garri49 [273]3 years ago
8 0

Answer:

9.51 Hz

Explanation:

f₀ = frequency of tuning fork and fundamental frequency of the pipe initially = 240 Hz

T₀ = Room temperature = 20 °C

v₀ = speed of sound at room temperature

speed of sound at room temperature is given as

v₀ = 331 + (0.6) T₀ = 331 + (0.6) (20) = 343 m/s

L₀ = Length of the pipe

For open pipe, we have  

v₀ = 2 L₀ f₀

343 = 2 (240) L₀

L₀ = 0.715 m

T = new temperature = 43.0°C

v = speed of sound at new temperature

speed of sound at new temperature is given as

v = 331 + (0.6) T = 331 + (0.6) (43) = 356.8 m/s

L = length of pipe = L₀ = 0.715 m

f = new fundamental frequency

For open pipe, we have  

v = 2 L f

356.8 = 2 (0.715) f

f = 249.51 Hz

Δf = Beat frequency

Beat frequency is given as

Δf = f - f₀

Δf = 249.51 - 240

Δf = 9.51 Hz

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one-third of its weight on Earth's surface

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6 0
4 years ago
Until he was in his seventies, Henri LaMothe excited audiences by belly-flopping from a height of 9 m into 32 cm. of water. Assu
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Answer:

823.46 kgm/s

Explanation:

At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.

So, mgh = 1/2mv²

From here, his velocity just as he reaches the surface of the water is

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His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,

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6 0
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Answer:

0.918 seconds

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t = v / g

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t = 0.918s

5 0
3 years ago
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