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3 years ago

7

A wave is traveling at a speed of 12m/s and its wavelength is 3 m calculate the wave frequency

1 answer:

White raven [17]3 years ago

6 0

Answer:

f = 4 Hz

Explanation:

Given that,

The speed of a wave, v = 12 m/s

The wavelength of a wave, $\lambda=3\ m$

We need to find the frequency of the wave. We know that the speed of wave is equal to the product of wavelength and frequency. So,

$v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{12}{3}\\\\f=4\ Hz$

So, the frequency of the wave is equal to 4 Hz.

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melomori [17]

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a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

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$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

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$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

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3 years ago