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4 years ago

What is the frequency of blue light that has a wavelength of 446 nm?

2 answers:

7 0

When I went through with the math, the answer I came upon was:
6.67 X 10^14 

Here is how I did it: First of all we need to know the equation. 

c=nu X lamda 
(speed of light) = (frequency)(wavelength) 
(3.0 X 10^8 m/s) = (frequency)(450nm) 

We want the answer in meters so we need to convert 450nm to meters. 
450nm= 4.5 X 10^ -7 m 
(3.0 X 10^8 m/s) = (frequency)(4.5 X 10^ -7 m) 

Divide the speed of light by the wavelength. 
(3.0 X 10^8m/s) / (4.5 X 10^ -7m) =6.67 X 10^ 14 per second or s- 

Answer: 6.67 X 10^14 s- hope this helps

BartSMP [9]4 years ago

3 0

From Electromagnetic Waves,
Velocity C = f *λ

Where C = Velocity in m/s. = 3 * 10^8 m/s.
f = Frequency in Hertz    ( /s)
   λ = Wavelength in metres.

Wavelength of blue light = 446 nm = 446 nano metres = 446 * 10^-9 m.

f = C / λ = ( 3 *10^8) / (446 * 10^-9) use your calculator.
   = 6.7265 * 10 ^ 14 Hertz.

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What type of device forms images through refraction?

Tcecarenko [31]

Answer:

A lens

Explanation:

A lens forms images when light passes Through it bending the rays of in the process.A phenomena called refraction and the speed of light changes in the process because it enters a medium since it's wavelength is reduced.

3 0

2 years ago

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago

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