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3 years ago

Comparing sound waves with different frequencies and wavelengths traveling through air,

Physics

1 answer:

aleksklad [387]3 years ago

4 0

Explanation:

The speed of sound wave only depends on the property of the medium like density and the bulk modulus of the medium particle. The speed of sound also depends on the temperature of the medium.

On comparing sound waves with different frequencies and wavelengths traveling through air, the speed of the wave doesn’t depend on the frequency or the wavelength. Hence, the correct option is (1).

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lakkis [162]

The right answer is “Atoms “

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3 years ago

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An ocean liner leaves New York City and travels 18.0o north of east for 155 km. How far east and how far north has it gone? In o

Monica [59]

I’m sorry if i took up a lot of space, hope this is a valid approximate answer

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1 year ago

Hi guys! pls help I asked this question almost 2 times and still didn't receive my answers.....Thanks in advance..

rjkz [21]

Answer:

1. Lateral inversion is a phenomenon in which left appears to be right and vice versa. It is due to direction that light follows when it strikes a reflecting surface, generally a mirror.

These are the letters which don't show lateral inversion A,H,O,T,U

2. USES OF CONCAVE MIRROR

They are used as shaving mirrors to see a larger image of the face.

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2 years ago

A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl

Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

$Q = C_w BH^{3/2}$

$H = (\dfrac{Q}{C_wB})^{2/3}$

$H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}$

$H = 0.371\ m$

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0

3 years ago

g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for

lbvjy [14]

Answer:

Explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

$a\text{sin}\theta=\frac{2N+1}{2}\lambda$

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

$0.110\times10^{-3}\text{sin}2.12=(\frac{2N+1}{2})582\times10^{-9}$

Solving the above equation we get

we N= 2.313≅ 2

4 0

3 years ago