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lesantik [10]
3 years ago
13

Radiation emitted from human skin reaches its peak at λ = 960 µm. what is the frequency of this radiation? planck's constant is

6.63 × 10−34 j · s. answer in units of hz. 003 (part 2 of 3) 10.0 points what type of electromagnetic waves are these? 1. gamma rays 2. visible light 3. radio waves
Physics
1 answer:
Artist 52 [7]3 years ago
6 0
1) The wavelength of the radiation emitted by the human skin is
\lambda=960 \mu m = 960 \cdot 10^{-6} m
the frequency of the radiation is related to the wavelength by
f= \frac{c}{\lambda}
where c=3 \cdot 10^8 m/s is the speed of light. Plugging numbers into the formula, we find the frequency of the radiation:
f= \frac{3 \cdot 10^8 m/s}{960 \cdot 10^{-6}m}=3.13 \cdot 10^{11} Hz

2) The frequency of this radiation is 313 GHz, and its wavelength 960 \mu m. If we look at the table of the electromagnetic spectrum
https://en.wikipedia.org/wiki/Electromagnetic_spectrum
We see that we are in the range of visible light (in particular, in the infrared range).
Therefore, the correct answer is <span>2. visible light .</span>
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Answer:

the pressure at B is 527psf

Explanation:

Angular velocity, ω = v / r

ω = 20 /1.5

= 13.333 rad/s

Flow equation from point A to B

P_A+rz_A-\frac{1}{2} Pr_A^2w^2=P_B+rz_B-\frac{1}{2} pr^2_Bw^2\\\\P_B = P_A + r(z_A-z_B)+\frac{1}{2} pw^2[(r_B^2)-(r_A)^2]\\\\P_B = [25 +(0.8+62.4)(0-1)+\frac{1}{2}(0.8\times1.94)\times(13.333)^2[2.5^2-1.5^2]  ]\\\\P_B = 25 - 49.92+551.79\\\\P_B = 526.87psf\\\approx527psf

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