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Elena L [17]
3 years ago
15

Determine whether the interference is constructive or destructive at each location indicated.

Physics
1 answer:
OlgaM077 [116]3 years ago
6 0

Answer:

A. Constructive

B. Destructive

C. Destructive

D. Constructive

Explanation:

Constructive interference takes place at locations along the path of two superposed waves where the waves are in phase such that a high or low point of one of the waves corresponds with a high or low point of the other wave which gives a resulting wave amplitude which is the sum of the amplitudes of the individual waves

Destructive interference takes place at locations along the path of two superposed waves where one wave is out of phase with the other wave such that a high or low point of one of the waves coincides with a low or high point of the other wave respectively thereby cancelling the effect of the other wave and giving a resulting wave that has an amplitude which is the difference in the amplitudes of the individual waves

Therefore;

At point A, the peak of each wave partially coincides resulting in constructive interference

At point B, the peak of the blue wave and the trough of the red wave partially coincides resulting in destructive interference

At point C, the through of the blue wave and the peak of the back wave partially coincides resulting in destructive interference

At point D, the trough of each wave partially coincides resulting in constructive interference.

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A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder
QveST [7]

Answer:

\Delta \theta = 47.57^{\circ} C

Explanation:

given,

moles of air compressed, n = 1.70 mol

initial temperature, T₁ = 390 K

Power supply by the compressor, P = 7.5 kW

Heat removed = 1.3 kW

Angular frequency of the compressor, f = 110 rpm = 110/60 = 1.833 rps.

Time of compression = time of the hay revolution

             =\dfrac{1}{2}\ T

             =\dfrac{1}{2}\times \dfrac{1}{f}

             =\dfrac{1}{2}\times \dfrac{1}{1.833}

             =0.273 s

Using first law of thermodynamics

U = Q - W

now,

\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t}- \dfrac{\Delta W}{\Delta t}

Power supplied \dfrac{\Delta W}{\Delta t} = 7.5 kW

heat removed \dfrac{\Delta Q}{\Delta t} = 1.3 kW

now,

\dfrac{\Delta U}{\Delta t} = 7.5 -1.3

\dfrac{\Delta U}{\Delta t} = 6.2 kW

we know,

\dfrac{\Delta U}{\Delta t}=\dfrac{nC_v\Delta \theta}{\Delta t}

 C_v for air = 5 cal/° mol

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now,

\Delta \theta = \dfrac{\Delta U}{\Deta t}\times \dfrac{\Delta t}{n C_v}

\Delta \theta = 6.2\times 10^3 \times \dfrac{0.273}{1.7\times 20.93}

\Delta \theta = 47.57^{\circ} C

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Answer:

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Explanation:

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We can set up the following equations with the information given:

\mathrm{(a)}\: 2.0\cdot 10^3=\frac{1}{2}\cdot 1000\cdot v^2, \\v=\fbox{$2\: \mathrm{m/s}$}

For part B, we have the same equation, but kinetic energy is now 2.0\cdot 10^5.

Therefore:

\mathrm{(b)}\: 2.0\cdot 10^5=\frac{1}{2}\cdot 1000\cdot v^2, \\v=\fbox{$20\: \mathrm{m/s}$}.

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