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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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How much power will it take to move a 10 kg mass at an acceleration of 2 m/s² a distance of 10 meters in 5 seconds?
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Answer:

100 Watts

Explanation:

<u>These equations are needed to work out the answer:</u>

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Sound waves have 4 basic properties including amplitude, wavelength and
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What are electromagnetic waves?​
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5 0
3 years ago
Read 2 more answers
Jet aircraft maintenance crews are required to wear protective earplugs. Members of a particular crew wear earplugs that reduce
san4es73 [151]

Answer:

the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB

Explanation:

Given the data in the question;

sound intensity reduced by the factor, m = 305

the sound intensity level experienced by the crew members wearing protective earplugs, L = 79 dB

Now, using the expression of sound intensity level;

L = 10log( I_0 )

where  I_0 is the intensity at L level

so we substitute

79 = 10log( I_0 )

I_0 = 10^{7.9

Now, expression for actual intensity;

I = mI_0

where I is the actual intensity

so we substitute

I = 305 × 10^{7.9

Next, we write the expression of sound intensity level for reduced intensity;

L' = 10log( I )

So we substitute

L' = 10log( 305 × 10^{7.9 )

L' = 10log( 24227011159.09058 )

L' = 103.8 dB

Therefore, the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB

8 0
3 years ago
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