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snow_lady [41]
3 years ago
6

If two waves pass a point every second, what is the frequency of the waves?

Physics
1 answer:
lisabon 2012 [21]3 years ago
5 0

T = 1/f

So frequency is 1/T (period)

So if every second two waves pass through a given point then I believe the frequency would be 2Hz which is 2 waves every second, or 2 waves per second.

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3 years ago
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What is the value of a conversion factor ratio?<br>A.1<br>B.3<br>C.10<br>D.12
Alecsey [184]
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4 0
4 years ago
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

                   = 5 x 4.186 J/mol°C  = 20.93 J/mol°C

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

the temperature change per compression stroke is equal to 47.57°C.

4 0
3 years ago
A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.95 m and whose unstretched length is 0.65 m. Next the
nexus9112 [7]

Answer:

Explanation:

spring constant of spring = mg / x

= .4 x 9.8 / ( .95 - .65 )

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energy stored in spring = 1/2 k x²

= .5 x 13.07 x ( 1.2 - .65 )²

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Let it goes x m beyond its equilibrium position

Total energy at initial point

= 1.976 + 1/2 m v²

= 1.976 + .5 x .4 x 1.6²

= 2.488 J

energy at final point

= mgh + 1/2 k x²

.4 x 9.8 x  ( .55 + x ) + .5 x 13.07 x² = 2.488

6.535 x² + 2.156 + 3.92 x = 2.488

6.535 x² + 3.92 x - .332 = 0

x = .075 m

7.5 cm

4 0
3 years ago
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ipn [44]
<span>The correct answer is: Oxygen

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8 0
3 years ago
Read 2 more answers
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