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USPshnik [31]
3 years ago
5

Two of the types of infrared light, ir-c and ir-a, are both components of sunlight. their wavelengths range from 3000 to 1,000,0

00 nm for ir-c and from 700 to 1400 nm for ir-
a. compare the energy of microwaves, ir-c, and ir-
a.
Physics
2 answers:
crimeas [40]3 years ago
7 0

The energy of electromagnetic waves can be calculated by using the following formula:

E=\frac{hc}{\lambda}

where:

h=6.63 \cdot 10^{-34} Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength of the wave

Substituting the various wavelengths in the formula, we find:

IR-C:

\lambda=3000 nm=3000\cdot 10^{-9} m -->

E=\frac{hc}{\lambda}=\frac{(6.63 \cdot 10^{-34})(3 \cdot 10^8)}{3000\cdot 10^{-9}}=6.63 \cdot 10^{-20} J

\lambda=1,000,000 nm=1\cdot 10^{-3} m -->

E=\frac{hc}{\lambda}=\frac{(6.63 \cdot 10^{-34})(3 \cdot 10^8)}{1\cdot 10^{-3}}=1.99 \cdot 10^{-22} J

IR-A:

\lambda=700 nm=700\cdot 10^{-9} m -->

E=\frac{hc}{\lambda}=\frac{(6.63 \cdot 10^{-34})(3 \cdot 10^8)}{700\cdot 10^{-9}}=2.84 \cdot 10^{-19} J

\lambda=1400 nm=1400\cdot 10^{-9} m -->

E=\frac{hc}{\lambda}=\frac{(6.63 \cdot 10^{-34})(3 \cdot 10^8)}{1400\cdot 10^{-9}}=1.42 \cdot 10^{-19} J

Therefore, we see that IR-A have higher energy than IR-C.


NikAS [45]3 years ago
6 0
The energy of a light wave is calculated using the formula
E = hc/λ
h is the Planck's constant
c is the speed of light
λ is the wavelength
For the ir-c, the range is
<span>6.63 x 10^-34 (3x10^8) / 3000 = 6.63 x 10 ^-29 J
</span>6.63 x 10^-34 (3x10^8) / 1000000 = 1.99 x 10^-31 J

For the ir-a, the range is
6.63 x 10^-34 (3x10^8) / 700 = 2.84 x 10^-28 J
6.63 x 10^-34 (3x10^8) / 1400 = 1.42 x 10^-28 J
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The gas particles leaving the car exhaust show the process of combustion.

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Combustion refers to a chemical process in which a substance reacts with oxygen to give off heat. The original substance is known as the fuel, whereas the source of oxygen is known as the oxidizer. The fuel can be present in solid, liquid, or gas, but for an airplane, the fuel is usually a liquid form. Combustion also called burning that is the basic chemical process of releasing energy from a fuel and air mixture. In an internal combustion engine, the ignition and combustion of the fuel occurs within the engine. The engine converts the energy from the combustion process to work.

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A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie
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The range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $\theta

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $\theta

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $\theta

U = Iπr^{2}B cos $\theta

For $\theta = 0° →

U(Max) = MB cos(0) = MB =  Iπr^{2}B = 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

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For $\theta = 90° →

U = MB cos (90) = 0

For $\theta = 180° →

U(Min) = MB cos(0) = - MB =  - Iπr^{2}B = - 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

- 375 π x 10^{-7}.

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

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