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3 years ago

What is the wavelength of an ekectromagnetic wave that has a frequency of 40,000 hz? the speed of light is 3x10^8 m/s?

Physics

1 answer:

ehidna [41]3 years ago

7 0

Solution
As,
speed=frequency × wavelengths
so,
wavelength=speed/frequency
=3×10^8/40000
=7500m

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Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

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

The shear stress is equal to:

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

b) If the railing edge is 2 ft, the Reynold number is:

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

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

What kind of energy is stored in a flashlight battery enabling it to function? electrical energy, chemical energy, mechanical en

Lady bird [3.3K]

Electrical Energy because the electrons in the battery travel from out one end of the battery through a circuit and back to the other end

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4 years ago

Read 2 more answers

The current in a hair dryer measures 11 amps. The resistance of the hair dryer is 15 ohms. What is the voltage? Your Answer: Que

lorasvet [3.4K]

V=IR
V=15x11
V=165ohms

I don’t quite remember the unit

Ohms law

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3 years ago

A dust particle with mass of 5.4×10−2 g and a charge of 2.3×10−6 C is in a region of space where the potential is given by V(x)=

user100 [1]

Answer:

1.6 m/s^2

Explanation:

Hello!

To calculate the acceleration we must know the electric field. The electric field and the potential are related by:

$E = -\frac{dV}{dx} =- 2.6(\frac{V}{m^{2}})x + 8.1(\frac{V}{m^{3}})x^{2}$

If the particle starts at 2.3m, the electric field is:

E = 36.869 V/m = 36.869 N/C

So, the force on the particle is:

F = q E = 2.3×10^−6 C * 36.869 N/C = 8.48 x 10^-5 N

And its acceleration is :

a = F/m = 8.48 x 10^-5 N / 5.4×10−5 kg = 1.57 m/s^2

Rounded to two significant figures:

1.6 m/s^2

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4 years ago

The ease with which the charge distribution in a molecule can be distorted by an external electrical field is called the _______

Trava [24]

It is called polarizability. A dipole moment is needed for the molecule or atom due to the distortion of the electron cloud. A dipole is a pair of electrical charges that have opposite signs and during a dipole moment two charges separate.

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3 years ago