The sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec
<h3>Calculating wavelength </h3>
From the question, we are to determine how many times longer is the first sound wave compared to the second sound water
Using the formula,
v = fλ
∴ λ = v/f
Where v is the velocity
f is the frequency
and λ is the wavelength
For the first wave
f = 20 waves/sec
Then,
λ₁ = v/20
For the second wave
f = 16,000 waves/sec
λ₂ = v/16000
Then,
The factor by which the first sound wave is longer than the second sound wave is
λ₁/ λ₂ = (v/20) ÷( v/16000)
= (v/20) × 16000/v)
= 16000/20
= 800
Hence, the sound wave with a <u>frequency of 20</u> waves/sec is 800 longer than the wavelength of a sound wave with a <u>frequency of 16,000</u> waves/sec
Learn more on Calculating wavelength here: brainly.com/question/16396485
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Answer:
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Step-by-step explanation:
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Answer:
gusion,layla,alucard,balmon,fanny,freya,chou,lapu lapu,cecelion,carmila,miya,nana,harith,
Step-by-step explanation:
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Answer:
i don't understand your question, but i'll try to the best of my ability.
the answer to that is:
-8.33333333333
Step-by-step explanation:
let me break this down into a way you might understand.
1: 3-1=2
2: 2 divided by 3=0.66666666666.
3: i don't know if the "< 6;n>", was meant to be in the equation, so i just added the rest of it up.
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Answer:
70.8
Step-by-step explanation:
77% is the same as 0.77 so
92 * 0.77 = 77% of 92
92 * 0.77 = 70.84
rounded to the nearest tenth
70.8