The third harmonic of a tube closed at one end is 735 Hz. If the speed of sound

Question

The third harmonic of a tube closed at one end is 735 Hz. If the speed of sound in air is 335 m/s, the length of the tube must be A) 11.6 cm B) 22.9cm C) 34.1 cm D) 45.7 cm E) 57.3 cm A string is connected to a tuning fork whose frequency is 80.0 Hz and is held under tension by 0.500 kg. The tuning fork causes the string to vibrate as shown. The mass per unit length for the string is A) 9.45 times 10^-4 kg/m D) 6.00 times 10^-3 kg/m B) 6.80 times 10^-3 kg/m E) 3.85 times 10^-2 kg/m C) 4.34 kg/m A mass of air is at pressure P and volume V at 1 degree C. When it is made to occupy half this volume at three times this pressure, its temperature becomes approximately. A) 411 degree C B) 0.67 degree C C) 1.5 degree C D) 6 degree C E) 411 K A 4-kg mass of metal of unknown specific heat at a temperature of 600 degree C is dropped into 0.5 kg of ice and 0.5 kg of water both at 0 degree C. With no heat losses to the surroundings, the equilibrium temperature of the mixture is 85 degree C. Calculate the specific heat of the metal. A) 0, 04 kcal/kg degree C D) 1.6 kcal/kg degree C B) 0.06 kcal/kg degree C E) 1.2 kcal/kg degree C C) 0.08 kcal/kg degree C Suppose you do 75 kJ of work on a system consisting of 10 kg of water by stirring it with a paddle wheel. During this process. 40 kcal of heat is removed. The change in the internal energy of the system is A) -35 kJ 13) -115 kJ C)-134 kJ D) -242 kJ E) -156 kJ

Explanation / Answer

6.

for third harmonic,

f = 3v/4L

or, L = 5v/4f = 3(335m/s)/4(735Hz)

or, L = 34.1 cm

So correct option is C)

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7.

Here wavelength is = 1.8m/2 = 0.9m

frequency is f = 80 Hz

So, speed of wave is v = f = (80 Hz)(0.9 m) = 72 m/s

Tension in the string is T = mg = (0.5 kg)(9.8 m/s2) = 4.9 N

Now, (T/µ)1/2 = v = 72 m/s, where µ is mass per unit length.

or, µ = (4.9 N)/(72 m/s)2

or, µ = 9.45X10-4 kg/m

So correct option is A)

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8.

Let the number of moles be n, then,

PV/T0 = (3P)(V/2)/T, where P, V' and T0 = 10C = 274 K, are initial pressure, volume and temperature

or, 1/T0 = 3/2T

or, T = (3/2)T0 = (3/2)(274 K)

or, T = 411 K

So, the correct option is E)

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9.

let the specific heat of metal be s, then

(4 kg)s(6000C - 850C) = (0.5 kg)(4187 J/kg-K)(850C) + (0.5 kg)(4187 J/kg-K)(850C) + (0.5 kg)(3.33X105 J/kg)

or, s = 253.6 J/kg-K = 0.06 kcal/kg0C

So the correct option is B)

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

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