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klemol [59]
3 years ago
5

Consider the following three statements: (i) For any electro-magnetic radiation, the product of the wavelength and the frequency

is a constant. (ii) If a source of light has a wavelength of 3.0 Å, its frequency is 1.0*1018 Hz. (iii) The speed of ultra-violet light is greater than the speed of microwave radiation. Which of these three statements is or are true?
Physics
2 answers:
kow [346]3 years ago
7 0

Answer:

Option (i) For any electro-magnetic radiation, the product of the wavelength and the frequency is a constant.

(ii) If a source of light has a wavelength of 3.0 Å, its frequency is 1.0*1018 Hz.

Explanation:

The wave equation is given as:

v = f\lambda

where f  = frequency

           λ = wavelength

Thus, the product of the wavelength and frequency produces a velocity of the wave not a constant.

The velocity of the wave depends of both properties.

(ii) The wavelength of the light changes depending on the source of the radiation. Thus, the frequency is not constant or universal for one form of a wave.

Therefore statements (i) and (ii) are false.

Scilla [17]3 years ago
3 0

Answer:

A and B

Explanation:

The relation between frequency and wavelength is shown below as:

c=frequency\times Wavelength

c is the speed of light having value 3\times 10^8\ m/s

Thus, the product of the wavelength and the frequency is constant and equal to 3\times 10^8\ m/s

<u>Option A is correct.</u>

Given, Frequency = 1\times 10^{18}\ Hz

Thus, Wavelength is:

Wavelength=\frac{c}{Frequency}

Wavelength=\frac{3\times 10^8}{1\times 10^{18}}\ m

Wavelength=3\times 10^{-10}\ m

Also, 1 m = 3\times 10^{-10} Å

So,

<u>Wavelength = 3.0 Å</u>

<u>Option B is correct.</u>

As stated above, the speed of electromagnetic radiation is constant. Hence, each radiation of the spectrum travels with same speed.

<u>Option C is incorrect.</u>

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None

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3 years ago
(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur
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Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

(b) The ratio of the intensities of the waves is 0.000535

Explanation:

Given;

density of gas, \rho _g = 2.27 kg/m³

density of liquid, \rho _l = 972 kg/m³

speed of sound in gas, C_g = 376 m/s

speed of sound in liquid, C_l = 1640 m/s

The of the sound wave is given by;

I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}

Where;

P_o is the pressure amplitude

P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535

3 0
3 years ago
In practice, if a voltmeter was connected across any combination of the terminals, the potential difference would be less than w
VladimirAG [237]

Answer:

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

Explanation:

A voltmeter is built by a galvanometer and a resistance in series, this set is connected in parallel to the resistance where the voltage is to be measured, therefore the voltage is divided between the voltmeter and the element to be measured, consequently the measured voltage It is less than the calculated one, since for them the resistance of the voltmeter is assumed infinite.

This difference is kept to a minimum because the resistance in transformers is a few tens of ohms and the resistance of modern voltmeters is of the order of MΩ.

8 0
3 years ago
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Artist 52 [7]
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Answer:

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Explanation:

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