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Elena-2011 [213]
3 years ago
5

With what type of electromagnetic radiation would you observe:

Physics
1 answer:
Viefleur [7K]3 years ago
5 0

Answer:

Green part of the visible spectrum.

X ray part of the electromagnetic spectrum.

Infrared part of the electromagnetic spectrum.

Explanation:

Wien's displacement law

\lambda_{max}=\frac{b}{T}

Where, b = Wien's displacement constant = 2.898×10⁻³ mK

T = Temperature in kelvin

\lambda_{max}=\frac{2.898\times 10^{-3}}{5800}\\\Rightarrow\lambda_{max}=0.499\times 10^{-6}=0.5\mu m=500\ nm

So, the wavelength would be of around the green part of the visible spectrum.

\lambda_{max}=\frac{2.898\times 10^{-3}}{1\times 10^6}\\\Rightarrow\lambda_{max}=2.898\times 10^{-9}=2.898\ nm

So, the wavelength would be of around the X ray part of the electromagnetic spectrum.

Human body temperature = 37°C = 37+273.15 = 310.15 K

\lambda_{max}=\frac{2.898\times 10^{-3}}{37+273.15}\\\Rightarrow\lambda_{max}=\frac{2.898\times 10^{-3}}{310.15}\\\Rightarrow\lambda_{max}=9.34\times 10^{-6}=934\ nm

So, the wavelength would be of around the Infrared part of the electromagnetic spectrum.

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Answer: it's under explanation

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Function Of The glottis​
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3 years ago
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The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

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3 years ago
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2+2=4

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A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back
elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=9.8 m/s^2

a.We have to find the force by which the box accelerates forward.

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Force by which box accelerates=62.286 N

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

ma=\mu mg

a=\mu g=9.8\times 0.202=1.9796 m/s^2

Hence, the truck can have maximum acceleration before the box slide=1.9796 m/s^2

3 0
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