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o-na [289]
3 years ago
15

A sound wave is determined to have a frequency of 1,000 Hz and a wavelength of 35 cm. What is the speed of this wave? If the sou

nd wave in the previous question is measured in air, what is the temperature of the air?
Physics
1 answer:
mars1129 [50]3 years ago
4 0

Answer:

350 m/s

31°C

Explanation:

Speed of sound is given as the product of frequency and wavelength

S=fw

Where s represent the speed in m/s, f is frequency and w is wavelength

Conversion

Taking 1m to be 100 cm then 35 cm will be 35/100=0.35m

Substituting 0.35 m for w and 1000 Hz for f then

S=1000*0.35=350 m/s

speed of sound (m/s) = 331.5 + 0.60 T(°C)

350=331.5+0.6T

T=30.833333333333

The temperature is approximately 31°C

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3 years ago
I don’t have a question because I posted a photoz
Scorpion4ik [409]

Part a:

Q_{1} = 56

Q_{2} = 60

Q_{3} = 63

     The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The Q_{2} is the overall medium, Q_{1} is the medium of the first half, and Q_{3} is the medium of the second half.

-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.

[] See attached

Part b:

The range is 7.

The interquartile range is the range of numbers between Q_{1} and Q_{3}. In other words, it is 50% of the data, directly in the middle.

This becomes 63 - 56 = 7

Part c:

79 is an outlier.

It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.

-> 63 + (7 + \frac{7}{2}) ≤ 79

-> 63 + 10.5 ≤ 79

-> 73.5 ≤ 79

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

7 0
2 years ago
If an astronaut has a mass of 80kg on earth, what is their mass on the moon?
stealth61 [152]

Answer:

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Explanation:

6 0
3 years ago
If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the
kodGreya [7K]

Answer:

a)

v = 4.048 *10^6 m/s

b)  

Angular frequency =  1.92 * 10^7

Explanation:

As we know

v =  \frac{qBr}{m}

q is the charge on the electron = 3.2 * 10^{-19} C

B is the magnetic field in Tesla = 0.4 T

r is the radius of the circle = 0.21 m

mass of the electrons = 6.64 * 10^{-27} Kg

a)

Substituting the given values in above equation, we get -

v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6m/s

b)  

Angular frequency =

\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7

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2 years ago
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Answer:

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Explanation:

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