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Svetradugi [14.3K]
4 years ago
10

High-frequency sound waves have a shorter (amplitude, pitch, wavelength) and a higher (amplitude, pitch, wavelength) than low-fr

equency sound waves.
Physics
2 answers:
Elanso [62]4 years ago
8 0

Answer: High-frequency sound waves have a shorter <u>wavelength </u>and a higher <u>pitch</u> than low-frequency sound waves.

Explanation:

For wave moving in a particular medium, its seed is constant. The wavelength of the wave is inversely proportional to the frequency. The pitch is the quality of sound which directly proportional to the frequency. Higher the frequency, higher is the pitch.

Thus, a high frequency sound wave would have shorter wavelength and higher pitch as compared to a low frequency sound waves.

I am Lyosha [343]4 years ago
5 0
The most appropriate answers are ...

high frequency sound has shorter wavelength !

a higher pitch than low frequency...!!

so , wavelength and pitch !!
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Listening to your favorite radio station involves which area of physics?
lukranit [14]

If I'm not mistaken the answer would be c, vibrations and wave phenomena, because when sound waves hit your ear they go through it to your brain and your mind translates these vibrations to sound. I hope my answer helped you out.

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3 years ago
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8 0
2 years ago
A thin film soap bubble (n=1.35) is floating in air. If the thickness of the bubble wall is 300nm, which of the following wavele
iren [92.7K]

Answer:

540 nm

Explanation:

According to the question,

The refractive index of the soap bubble, n=1.35.

The thickness of the soap bubble wall is, t=300 nm.

Now, for constructive interference of soap bubble.

2nt=(m+\frac{1}{2})\lambda.

Now for first order m=1.

Therfore,

\lambda =\frac{4}{3} tn

Substitute all the variables in the above equation.

\lambda =\frac{4}{3} (1.35)(300 nm).

Therefore,

\lambda =540 nm.

Therefore the visible light wavelength which is strongly reflected is 540 nm.

6 0
4 years ago
An aluminum cup contains 225 g of water and a 40-g copper stirrer, all at 27°C. A 470-g sample of silver at an initial temperat
vagabundo [1.1K]

Answer:

Mal = 0.232 kg = 232 g

Explanation:

mass of water (Mw) = 225 g = 0.225 kg

mass of copper stirrer (Mcu) = 40 g = 0.04 kg

initial temperature of water (Tw) = 27 degrees

mass of silver (Mag) = 470 g = 0.47 kg

initial temperature of silver (Ts) = 85 degrees

final temperature of the mixture (T) = 32 degrees

find the mass of the aluminum cup (Mal)

applying the conservation of energy

((Mal.cAl) + (Mw.cW) + (Mcu.cCu))(ΔTw) = (Mag.cAg)(ΔTag)

we require the specific heat capacities of water (cW) , aluminium (cAl) , copper (cCu) and silver (cAg) which are as follows

water (cW) =4186 J/kg.K

aluminium (cAl) = 900 J/kg.K

copper (cCu) = 387 J/kg.K

silver (cAg) = 234 J/kg.K

now we can put in our values to get the mass of the Aluminium cup (Mag)

((Mal.900) + (0.225 x 4186) + (0.04 x 387))(32-27) = (0.47 x 234)(85-32)

(900 . Mal + 957.33) x 5 = 5828.94

900 .Mal + 957.33 = 1165.79

900.Mal = 1165.79 - 957.33 = 208.5

Mal = 0.232 kg = 232 g

8 0
4 years ago
How much positive electric charge is in 10 moles of carbon? NA=6.022×1023mol−1, e=1.60×10−19C.
levacccp [35]

Answer:

4.75 x 10^-25 kg

Explanation:

5 0
3 years ago
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