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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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Describe the frequency and wavelength range of radio waves
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Radio waves are a type of electromagnetic radiation with wavelengths in the electromagnetic spectrum longer than infrared light. They have frequencies from 300 GHz to as low as 3 kHz, and corresponding wavelengths from 1 millimeter to 100 kilometers.

Explanation:

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3 years ago
The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5
nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

G\frac{mM}{R^2}=m\frac{v^2}{R}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

m=11,100 kg is the mass of the telescope

M=5.97\cdot 10^{24} kg is the mass of the Earth

R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s

6 0
3 years ago
The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A
love history [14]

Answer:

Part a)

EMF = 14 \times 10^{-3} V

Part b)

EMF = 15.67 \times 10^{-3} V

Explanation:

As we know that magnetic flux through the loop is given as

\phi = B.A

now we have

\phi = B\pi r^2

now rate of change in flux is given as

\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}

now we know that

A = \pi r^2

0.285 = \pi r^2

r = 0.30 m

Now plug in all data

EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)

EMF = 14 \times 10^{-3} V

Part b)

Now the radius of the loop after t = 1 s

r_1 = r_0 + \frac{dr}{dt}

r_1 = 0.30 + 0.037

r_1 = 0.337 m

Now plug in data in above equation

EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)

EMF = 15.67 \times 10^{-3} V

5 0
3 years ago
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Whenever an object is falling, its potential energy
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7 0
3 years ago
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