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Hoochie [10]
3 years ago
12

We spend our lives in visible light without risks to our health. Yet exposure to UV waves in sunlight can cause skin cancer. Wha

t is the BEST explanation for this?
A) UV waves are longer and carry more energy.
B) UV waves are higher frequency and carry more energy.
C) UV waves have a lower frequency and carry more energy.
D) UV waves travel faster than visible light and have more energy.
Physics
2 answers:
dedylja [7]3 years ago
6 0
<span>UV waves are higher frequency and carry more energy causing exposure to UV waves in sunlight to skin cancer. This is a type of electromagnetic radiation similar to the waves of the radio, infrared, x-rays and gamma. All of which comes from the sun however cannot be seen by a human eye. People use these rays for tanning however if it is overdone, it may cause skin problems.</span>
Trava [24]3 years ago
3 0
The correct answer is
<span>B) UV waves are higher frequency and carry more energy. 

In fact, UV waves have higher frequency than visible light. For comparison, visible light has frequency in the range 430-770 THz (</span>10^{12} Hz)<span>, while ultraviolets (UV) have frequency higher than these values (at the order of 1 PHz, </span>1\cdot 10^{15}Hz).

The energy of electromagnetic radiation is proportional to its frequency, according to the equation
E=hf
where h is the Planck constant and f is the frequency. We see that the higher the frequency, the greater the energy, so UV waves carry more energy than visible light.
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Answer:

 t  = 16.5s

Explanation:

Given parameters:

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Initial velocity  = 0m/s

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Unknown:

Time taken  = ?

Solution:

To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.

So;

        Acceleration  = \frac{v  - u }{t}  

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u is the initial velocity

t is the time taken

  So;

       3.1  = \frac{51 - 0}{t}  

    3.1t  = 51

       t  = 16.5s

8 0
3 years ago
What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan
kramer
I think the question should be the below:

<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:

 <span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>

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<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>
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Answer:

<u>Example of Newton's III law</u>

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