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Tema [17]
3 years ago
15

When you go out into the sun, the UV light can give you a tan. UV light has a frequency of 7.89 x 1014 Hz. What is the wavelengt

h of UV light?
Please use 3 x 108 m/s for the speed of light (c).
Physics
1 answer:
Olin [163]3 years ago
6 0

Answer:

Explanation:

Formula and givens

  • λ = c / f
  • λ is the wavelength
  • c = the speed of light
  • f = the frequency
  • c = 3*10^8
  • f = 7.89 * 10^14

λ = ?

Solution

λ = 3*10^8 / 7.89*10^14

λ = 3*10^8/7.89*10^14

λ = 2.36 * 10^7

λ = 236 nanometers. What you use as your solution depends on what what you have been taught.

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A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with
Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D =  1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7}  m^{2}\\

Intensity

I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }

Amplitude of Electric Field

E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C

Amplitude of Magnetic Field

B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{  3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T

5 0
4 years ago
How is a fireplace an example of both convection and radiation?
slega [8]
A fire spreads by transferring heat in 3 ways, Raditation, convection, and Conduction :)
5 0
3 years ago
How would you find the speed of a ball thrown in to the air- it takes 6 seconds to get tho the top. The height is not given.
Alecsey [184]
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7 0
3 years ago
An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its l
Murrr4er [49]

Answer:

v = 25.45 m/s

Explanation:

In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:

h_{max}=\frac{v_o^2}{g}   (1)

vo: initial speed of the object = 18 m/s

g: gravitational acceleration = 9.8 m/s²

Furthermore you use the following formula for the final speed of the object:

v^2=v_o^2-2gh       (2)

h: height

You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:

v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2

Then, you solve the previous result for vo:

v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}

The initial speed of the object was 25.45 m/s

3 0
3 years ago
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