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PtichkaEL [24]
3 years ago
15

The sound produced by the loudspeaker in the drawing has a frequency of 11999 Hz and arrives at the microphone via two different

paths. The sound travels through the left tube LXM, which has a fixed length. Simultaneously, the sound travels through the right tube LYM, the length of which can be changed by moving the sliding section. At M, the sound waves coming from the two paths interfere. As the length of the path LYM is changed, the sound loudness detected by the microphone changes. When the sliding section is pulled out by 0.030 m, the loudness changes from a maximum to a minimum. Find the speed at which sound travels through the gas in the tube.
Physics
1 answer:
const2013 [10]3 years ago
3 0

The speed at which sound travels through the gas in the tube is 719.94m/s

<u>Explanation:</u>

Given:

Frequency, f = 11999Hz

Wavelength, λ = 0.03m

Velocity, v = ?

Sound speed in the tube is calculated by multiplying the frequency v by the wavelength λ.

As the sound loudness changed from a maximum to a minimum, then we know the sound interference in the case changed from constructive interference (the two sound waves are in phase, i.e. peaks are in a line with peaks and so the troughs), to a destructive interference (peaks coinciding with troughs). The least distance change required to cause such a change is a half wavelength distance, so:

λ/2 = 0.03/2

 λ  = 0.06m

We know,

v = λf

v = 0.06 X 11999Hz

v = 719.94m/s

Therefore, the speed at which sound travels through the gas in the tube is 719.94m/s

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An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
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a) The work done by the applied force is 1500 joules.

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Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force (W), measured in joules, is defined by the following expression:

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b) At first we need to calculate the net acceleration of the object (a), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

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\frac{1}{2}\cdot a \cdot t^{2} =  \Delta x-v_{o}\cdot t

a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}

If we know that \Delta x = 100\,m, v_{o} = 0\,\frac{m}{s} and t = 10\,s, then the net acceleration experimented by the object is:

a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}

a = 2\,\frac{m}{s^{2}}

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

\Sigma F = F - f = m\cdot a (3)

Where:

F - External force exerted on the object, measured in newtons.

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If we know that F = 15\,N, m = 6\,kg and a = 2\,\frac{m}{s^{2}}, the kinetic friction force is:

f = F-m\cdot a

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\Delta K = \Delta W (5)

If we know that \Delta W = 1200\,J, then the change in translational kinetic energy is:

\Delta K = 1200\,J

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

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