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3 years ago

Is noise energy a kind of energy or not

Physics

1 answer:

Volgvan3 years ago

4 0

Yes. sound waves are produced by energy.

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Find the first three harmonics of a string of linear mass density 2. 00 g/m and length 0. 600 m when the tension in it is 50. 0

gavmur [86]

The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

<h3>Velocity of the wave</h3>

The velocity of the wave is calculated as follows;

v = √T/μ

where;

T is tension
μ is mass per unit length = 2 g/m = 0.002 kg/m

v = √(50/0.002)

v = 158.1 m/s

<h3>First harmonic or fundamental frequency of the wave</h3>

f₀ = v/λ

where;

λ is the wavelength = 2L

f₀ = v/2L

f₀ = 158.1/(2 x 0.6)

f₀ = 131.8 Hz

<h3>Second harmonic of the wave</h3>

f₁ = 2f₀

f₁ = 2(131.8 Hz)

f₁ = 263.6 Hz

<h3>Third harmonic of the wave</h3>

f₂ = 3f₀

f₂ = 3(131.8 Hz)

f₂ = 395.4 Hz

Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

Learn more about harmonics here: brainly.com/question/4290297

#SPJ1

6 0

1 year ago

What is weightlessness ?

VMariaS [17]

Answer:

The state of being free from the effects of gravity.

7 0

3 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

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enyata [817]

Answer:

I think its distance

Explanation:

when measuring how far a p.o art u can use mm

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Read the given list of organisms.

Nadya [2.5K]

They are all herbivores

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