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3 years ago

What sound frequency would the human ear not be able to detect?

Physics

2 answers:

egoroff_w [7]3 years ago

7 0

It would be A. 10 Hertz
A average human can hear between only 20Hz - 20,000Hz

katrin2010 [14]3 years ago

3 0

Its 10 hertz i took the test

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In music, the note G above middle C has a frequency of about 392 hertz. If the speed of sound in the air is 340 m/s, what is the

morpeh [17]

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0.87 meters

Explanation:

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2 years ago

How to measure the external diameter of a sphere

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3 years ago

Explain how tidal forces are causing Earth to slow down.

dybincka [34]

Answer: The tidal forces exerted by the moon are directly associated with the earth's rotation. Due to the strong gravitational pull of the moon, the tidal bulging appears on both the sides on earth and these are region of high tide, and there is gradual rise and fall of sea level.

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3 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago

Which statement correctly describes mass-energy equivalence? All energy in the universe will be converted to an equivalent amoun

olchik [2.2K]

The statement 'all energy in the universe is a result of mass being converted into energy' correctly describes mass-energy equivalence.

<h3>What is mass-energy equivalence?</h3>

The expression mass-energy equivalence refers to the proportion of matter that can be converted into energy in the universe.

This mass-energy equivalence is an outcome of process of converting mass into energy.

In conclusion, the statement 'all energy in the universe is a result of mass being converted into energy' correctly describes mass-energy equivalence.

Learn more about mass-energy equivalence here:

brainly.com/question/3171044

#SPJ1

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2 years ago