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leonid [27]
3 years ago
8

A vertical tube one meter long is open at the top. It is filled with 75 cm of water. If the velocity of sound is 344 m/s, what w

ill the fundamental resonant frequency be (in Hz)?
Physics
1 answer:
alukav5142 [94]3 years ago
7 0

Answer:

344Hz

Explanation:

First we need to more that the frequency will inky resonate in the portion not containing water. If the total length is 1m and the filled with water up to 75cm, the length of the air column will be 100cm - 75cm = 25cm

Fundamental frequency of a closed pipe fo = V/4L

V is the velocity of sound in air

L is the length of the air column

Given V = 344m/s

L = 25cm = 0.25m

fundamental resonant frequency = 344/4(0.25)

= 344/1

= 344Hz

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pashok25 [27]

Answer:

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m/(m/s) = m*s/m, cancels out giving s as a unit.

<h2><u>Therefore the answer is 232.56 seconds</u></h2>

6 0
2 years ago
The visible light portion of the electromagnetic spectrum is often subdivided into the colors of red, orange, yellow, green, blu
Ugo [173]

Light at the red end of the visible portion has the least energy, lowest frequency, same speed, and longer wavelength compared to the violet end.

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The part of the electromagnetic spectrum that can be seen by the human eyes is the visible spectrum. The light waves with the wavelengths of  380 to 740 nm can be sen by the human eyes. Light at the red end of the visible portion has the least energy, lowest frequency, same speed, and longer wavelength compared to the violet end.

5 0
3 years ago
Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = m_1=6.50 kg- 1.55 kg=4.95 kg

Velocity achieved by squid = v_1=2.60 m/s

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Mass of water = m_2=1.55 kg

Velocity by which water was released by squid = v_2

Momentum gained by water but in opposite direction = P'=m_2v_2

P = P'

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v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E  =\frac{1}{2}m_1v_1^{2}

Kinetic energy of water = K.E' = \frac{1}{2}m_2v_2^{2}

Total kinetic energy created by this maneuver:

K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}

=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules

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