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brilliants [131]
2 years ago
6

in a lab, resonance tubes are used to determine experimentally the speed of sound. using the data given, evaluate the best appro

ximation for the speed of sound
Physics
1 answer:
jek_recluse [69]2 years ago
3 0

The approximate speed of sound is about 761 mph or 335.28 meters/second.

<h3>What's the approximate speed of sound?</h3>

If we consider the normal atmosphere at sea level, the speed of sound is about 761 mph or 335.28 meters/second. The speed of sound is different in different mediums such as solid and liquid.

So we can conclude that the approximate speed of sound is about 761 mph or 335.28 meters/second.

Learn more about speed here: brainly.com/question/4931057

#SPJ1

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VARVARA [1.3K]

Answer:

O D.

Explanation:

Physics has an aspect that deals with the study of energy

8 0
3 years ago
What type of heat does not require matter?
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It would be Thermal Radiation
4 0
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What is the formula to calculate the thickness of a test tube?​
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Having the inside dimensions (ID) and the outside dimensions (OD) will allow you to figure out the wall thickness on tubing. You would need to subtract the ID from the OD and then divide by two. This number is the wall thickness.

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3 years ago
The cube with 2.00m wide and 2.00m long and 2.00m high has a weight of 960.00 N what pressure does it exert
andre [41]
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3 0
3 years ago
A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
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