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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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Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t
Nataliya [291]

Answer:

 Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

                                     I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

                                     Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

                                     Q = cos^-1 ( sqrt (0.464))

                                     Q = cos^-1 (0.6811754546)

                                     Q = 47.06 degrees

                                   

8 0
4 years ago
What is the frequency of an ocean wave that is traveling at a speed of 45 m/s if it has a wavelength of 3 meters.
vaieri [72.5K]

Answer:

Frequency, f = 15 Hz      

Explanation:

We have,

Speed of an ocean wave is 45 m/s

Wavelength of a wave is 3 m

It is required to find the frequency of an ocean wave.

Speed of a wave, v=f\lambda, f = frequency of ocean wave

f=\dfrac{v}{\lambda}\\\\f=\dfrac{45}{3}\\\\f=15\ Hz

So, the frequency of an ocean wave is 15 Hz.

5 0
3 years ago
Which answer explains why plate B is moving underneath plate A
blsea [12.9K]

Answer:

A.

Explanation:

Earth is composed of different layers and one layer moves over another due to differences in the densities.

According to the physics of density, a substance having less density floats over a higher density substance. The oceanic crust has more density than the continental crust that is why continental crust float over oceanic crust.

So in the given example, plate B is moving below the plate A, it means plate B is more dense than plate A because plate B is composed of oceanic crust . <u>For example : continents float over the asthenosphere (a layer below the lithosphere).</u>

Hence, the correct answer is "A ".

4 0
3 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh
S_A_V [24]

Answer:

V₀ = 5.47 m/s

Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

V₀² = 3.04 m/(0.10126 s²/m)

V₀ = √30.02 m²/s²

<u>V₀ = 5.47 m/s</u>

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No...........................................................................Work for them urself

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