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Sunny_sXe [5.5K]
3 years ago
14

A wave is traveling at a speed of 15 m/s and it's wavelength is 5 m. Calculate the waves frequency

Physics
1 answer:
Anvisha [2.4K]3 years ago
3 0

Answer:

The frequency of this wave is 3\; \rm Hz.

Explanation:

The frequency f of a wave is the number of wavelengths that this wave covers in unit time (typically a second.)

The wave in this question travels at v = 15\; \rm m \cdot s^{-1}. In other words, this wave covers 15\; \rm m in unit time (a second.) How many wavelengths \lambda would that 15\; \rm m\; correspond to?

The question states that the wavelength of this wave is \lambda = 5\; \rm m. Therefore, there would be 15 / 5 = 3 wavelengths in the 15\; \rm m span that this wave covered in the unit time of one second (1\; \rm s.) Hence, the frequency of this wave would be 3\; \rm s^{-1} (three per second,) which is equivalent to 3\; \rm Hz (three Hertzs.)

In general, the frequency f of a wave with speed v and wavelength \lambda would be:

\displaystyle f = \frac{v}{\lambda}.

For the wave in this question:

\begin{aligned}f &= \frac{v}{\lambda} \\ &= \frac{15\; \rm m \cdot s^{-1}}{3\; \rm s} = 3\; \rm s^{-1} = 3\; \rm Hz\end{aligned}.

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Answer:

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Parametric equations for given line are

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Explanation:

The vector equation of the line is given by

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r₀ = (7, -8, 3)

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At these points the vector equation for this line is:

\overrightarrow{r}=\overrightarrow{r_{o}}+t\overrightarrow{v}\\\overrightarrow{r}=+t

Parametric equations for given line are

                                       x=7+t\\y=-8+6t\\z=3-13t

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Which force stops the car from moving?
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An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s
ad-work [718]

Answer:

a

 \theta  =  0.0022 rad

b

 I  =  0.000304 I_o

Explanation:

From the question we are told that  

   The  wavelength of the light is \lambda  = 550 \ nm  =  550 *10^{-9} \ m

    The  distance of the slit separation is  d = 0.500 \ mm = 5.0 *10^{-4} \ m

 

Generally the condition for two slit interference  is  

     dsin \theta =  m \lambda

Where m is the order which is given from the question as  m = 2

=>    \theta  =  sin ^{-1} [\frac{m \lambda}{d} ]

 substituting values  

      \theta  =  0.0022 rad

Now on the second question  

   The distance of separation of the slit is  

       d =  0.300 \ mm  =  3.0 *10^{-4} \ m

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      I  =  I_o  [\frac{sin \beta}{\beta} ]^2

Where  \beta is mathematically evaluated as

       \beta  =  \frac{\pi *  d  *  sin(\theta )}{\lambda }

  substituting values

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So the intensity is  

    I  =  I_o  [\frac{sin (0.06581)}{0.06581} ]^2

   I  =  0.000304 I_o

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3 years ago
A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be  in the opposite direction

Formula for the change in specific entropy can be calculated as:

s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

where;

s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

We obtain the specific properties of air at temperature at T_1 = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

Now replacing our values  into equation (2) from above; we have;

s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0
3 years ago
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