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3 years ago

Why echo sound cannot be heard in small room

1 answer:

7 0

They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.

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There is little vertical air movement in the __ because of the cold, heavy air at the bottom of the layer.

pogonyaev

The correct answer is the Mesosphere. <span>Mesosphere </span><span>is the layer of the Earth's atmosphere that is directly above the stratopause and directly below the mesopause.</span>

4 0

3 years ago

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A ray of light strikes a mirror. The angle formed between the incident ray and the reflected ray measures 64 º. What are the mea

julia-pushkina [17]

Answer:

Draw line perpendicular to mirror

Angle of incidence = angle between incident ray and perpendicular

Angle of reflection = angle between reflected ray and perpendicular

32 + 32 = 64

Angle of incidence = angle of reflection = 32 deg

6 0

2 years ago

A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

Minnie sota hits the end of a bar 1. 2 m long with a hammer. Sketch the standing wave on the bar for the following situations. T

ella [17]

(a) The wavelength of the wave for the fundamental mode is 2.4 m.

(b) The fundamental frequency of the wave is 2,708.33 Hz.

<h3>Wavelength of the wave for fundamental mode</h3>

The wavelength of the wave for the fundamental mode is calculated as follows;

Node to Node, N → N = λ/2

L = λ/2

λ = 2L

λ = 2 x 1.2

λ = 2.4 m

<h3>Fundamental frequency of the wave</h3>

The fundamental frequency of the wave is calculated as follows;

f = v/λ

f₀ = v/2L

f₀ = (6500)/(2 x 1.2)

f₀ = 2,708.33 Hz

The diagram of the wave for the fundamental mode is in the image uploaded.

Learn more about wavelength here: brainly.com/question/10728818

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2 years ago

How do you measure mass? Need now!

Yakvenalex [24]

With a ruler and a calculator

7 0

4 years ago

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