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maria [59]
3 years ago
14

How can a person detect infrared rays without an instrument?

Physics
2 answers:
just olya [345]3 years ago
8 0

Answer:

Explanation:

The infrared radiations are detected by their heating effects. they cause heating effects.

natita [175]3 years ago
7 0
So we want to know how can we detect infrared rays without an instrument. Infrared rays or heat, are a part of electromagnetic spectrum. We have specialized nerve cells in our skin called thermoreceptors that can detect differences in temperature that are produced by infrared part of EM spectrum.
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What are the units of Mass and Weight in the English System? Explain.
Andru [333]

Answer:

I tried...

Explanation:

Assuming the English system of mass and weight is the same as the Imperial system, the measurements are:

• 1 pound (lb) = 16 ounces (oz)

• 2,000 pounds (lbs) = 1 ton (T)

I think that's everything?

8 0
2 years ago
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A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel
erastovalidia [21]

Answer:

D.

Explanation:

To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by

2t = (m+\frac{1}{2})\frac{\lambda}{n}

Where,

t = thickness

\lambda=Wavelenght

m= is an integer

n= film/refractive index

We use this equaton because phase change is only present for gasoline air interface, but not at the gasoline-water interface. <em>The minimum t only would be when the value of m=0 then</em>

2nt = \frac{\lambda}{2}

t = \frac{560nm}{4*1-4}

t = 100nm

Therefore the correct answer is D. The minimum thickness of the film to see ab right reflection is 100nm

4 0
3 years ago
A guitar string has a linear density of 8.30 ✕ 10−4 kg/m and a length of 0.660 m. the tension in the string is 56.7 n. when the
Sedbober [7]
Ans: Beat Frequency = 1.97Hz

Explanation:
The fundamental frequency on a vibrating string is 

f =   \sqrt{ \frac{T}{4mL} }<span>  -- (A)</span>

<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>


Plug in the values in Equation (A)

<span>so </span>f = \sqrt{ \frac{56.7}{4*5.48*10^{-4}*0.66} }<span> = 197.97Hz </span>

<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
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