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loris [4]
3 years ago
12

The higher an electromagnetic wave's frequency, the lower its

Physics
2 answers:
Pepsi [2]3 years ago
7 0

Answer:

The higher an electromagnetic wave's frequency, the lower its wavelength

~batmans wife dun dun dun...

tensa zangetsu [6.8K]3 years ago
3 0

Answer:

The higher an electromagnetic wave's frequency, the lower its wavelength.

Explanation:

The frequency of a electromagnetic wave is given by the following relation :

c=f\times \lambda

Where

c = speed of light

f = frequency of a wave

\lambda = wavelength

f=\dfrac{c}{\lambda}

It is clear that the frequency of a wave has inverse relationship with its wavelength. So, higher the electromagnetic wave's frequency, the lower its wavelength.

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Which factor affects elastic potential energy but not gravitational potential energy?
julsineya [31]

Explanation:

The energy stored in an elastic objects as a result of deformation is called elastic potential energy. The energy stored in a spring is given by :

E=\dfrac{1}{2}kx^2

Where

k = spring constant

x = compression or stretching in an spring

While gravitational potential energy is given by :

PE = mgh

where

m = mass

g = acceleration due to gravity

h = height from ground

So, the factor affecting elastic potential energy but not gravitational potential energy is " spring constant ".

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4 years ago
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Describe two ways to change the frictional force between two solid surfaces.
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4 years ago
The wavelength of red helium-neon laser light in air is 632.8 nm.
e-lub [12.9K]

(a) 4.74\cdot 10^14 Hz

The frequency of an electromagnetic wave is given by:

f=\frac{c}{\lambda}

where

c=3.0\cdot 10^8 m/s is the speed of the wave in a vacuum (speed of light)

\lambda is the wavelength

In this problem, we have laser light with wavelength

\lambda=632.8 nm=6.33\cdot 10^{-7} m. Substituting into the formula, we find its frequency:

f=\frac{3.0\cdot 10^8 m/s}{6.33\cdot 10^{-7} m}=4.74\cdot 10^14 Hz

(b) 427.6 nm

The wavelength of an electromagnetic wave in a medium is given by:

\lambda=\frac{\lambda_0}{n}

where

\lambda_0 is the original wavelength in a vacuum (approximately equal to that in air)

n is the index of refraction of the medium

In this problem, we have

\lambda_0=632.8 nm

n = 1.48 (index of refraction of glass)

Substituting into the formula,

\lambda=\frac{632.8 nm}{1.48}=427.6 nm

(c) 2.03\cdot 10^8 m/s

The speed of an electromagnetic wave in a medium is

v=\frac{c}{n}

where c is the speed of light in a vacuum and n is the refractive index of the medium.

Since in this problem n=1.48, we find

v=\frac{3\cdot 10^8 m/s}{1.48}=2.03\cdot 10^8 m/s

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4 years ago
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The answer is  Al(NO3)3
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