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3 years ago

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Infrared radiation has frequencies from 3.0×1011 to 3.0×1014 Hz, whereas the frequency region for radio wave radiation is 3.0×10

5 to 3.0×107 Hz. We can say that: 1. The speed of infrared radiation is radio wave radiation. 2. The wavelength of infrared radiation is radio wave radiation.

1 answer:

expeople1 [14]3 years ago

4 0

Answer:

1. The speed of infrared radiation is greater than radio wave radiation

2. The wavelength of infrared radiation is less than the radio wave radiation

Explanation:

Given:

The radio wave radiation N₁ is 3x10⁵ to 3x10⁷ Hz and the infrared radiation N₂ is 3x10¹¹ to 3x10¹⁴ Hz. The speed is:

$\frac{velocity_{2} }{velocity_{1} } =\frac{3x10^{11} }{3x10^{5} } to\frac{3x10^{14} }{3x10^{7} } =1x10^{6} to1x10^{7}$

1. The speed of infrared radiation is greater than radio wave radiation

2. The wavelength is:

λ = C/N

where more C, less λ

Then, the wavelength of infrared radiation is less than the radio wave radiation

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If a ball is rolled down a frictionless inclined plane from a height of 3 feet, to what height would it roll up to on the other

mojhsa [17]

Answer:

It will roll up to its initial height from which it rolled from on the first hill

Explanation:

The given information are;

The height from which the ball is rolled = 3 feet[

The equation of the potential energy of the ball as it rolled down the frictionless inclined plane is given by the following equation;

m × g × h₁ = 1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r]

Where;

m = The mass of the ball

h = The from which the ball is rolled down

v = The velocity of the ball

r = The radius of the ball

Therefore, all the potential energy, m × g × h, is converted into kinetic energy when the ball reached the ground

Similarly, as the ball rolls up on the other side, by the principle of energy conservation, all the kinetic energy is transformed into potential energy as follows

1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r] = m × g × h₂, therefore, we have;

1/2 × m × v² + 1/2×[2/5× m × r²) × [v/r] = m × g × h₁ = m × g × h₂

Therefore since m × g × h₁ = m × g × h₂, then h₁ = h₂, the ball rises to the same height it started rolling from on the first hill.

3 0

3 years ago

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

4 years ago

an oscilloscope has a time/div setting of 50ms and there are 10 divisions on the time scale. a sine wave on the oscilloscope dis

mamaluj [8]

Answer:

The frequency is f = 6Hz

Explanation:

Given

The time division setting is $d_t =50ms = 50*10^{-3}s$

The number of division is $n= 10$

The total time for 10 division can be mathematically obtained as

$T_{total} = 10 *50*10^{-3}$

$=500 *10^{-3}s$

From the question the time for 3 cycle is

$T_{total} = 500*10^{-3}s$

Then the for one cycle which equivalent to the Period $(T)$ $= \frac{T_{total}}{3}$

$= \frac{500 *10^{-3}}{3}$

The frequency is generally given as $f = \frac{1}{period} = \frac{1}{T}$

Now substituting values we have

$f = \frac{3}{500 *10^{-3}}$

$= 6Hz$

4 0

4 years ago

An object initially at rest accelerates at 5 meters per second2 until it attains a speed of 30 meters per second. What distance

Tcecarenko [31]

Answer: 90m

Explanation:

Use Equation for distance:

S=a*t²/2

use eqation for acceleraton a=(Vf-Vs)/t

Vs- starting speed

Vf - final speed

a-accelaration

---------------------------------

a=5m/s²

Vs=0m/s

Vf=30m/s

use

a=(Vf-Vs)/t

to find time

t=(Vf-Vs)/a

t=30m/s/5m/s²

t=6s

Now calculate distance that object travel using:

S=(a*t²)72

s=(5m/s²*(6s)²)/2

S=90m

4 0

3 years ago

Read 2 more answers

The energy of the electron in a hydrogen atom can be calculated from the Bohr formula: In this equation stands for the Rydberg e

labwork [276]

Answer:

Wavelength, $\lambda=657\ nm$

Explanation:

The energy of the electron in a hydrogen atom can be calculated from the Bohr formula as :

$E=\dfrac{-R}{n^2}$ .............(1)

Where

R is the Rydberg constant

n is the number of orbit

We need to find the wavelength of the line in the absorption line spectrum of hydrogen caused by the transition of the electron from an orbital with to an orbital with n₁ = 2 to an orbital with n₂ = 3.

Equation (1) can be re framed as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

$\dfrac{1}{\lambda}=1.096\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{3^2})$

$\lambda=6.569\times 10^{-7}\ m$

or

$\lambda=657\ nm$

So, the the wavelength of the line in the absorption line spectrum is 657 nm. Hence, this is the required solution.

3 0

3 years ago