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:
#1 is -9
#2 is 3
Step-by-step explanation:
-9x3 is -27 then add 10 to that you get -17
5x3 is 15 then subtract 1 and you get 14
Answer: C
<u>Step-by-step explanation:</u>
f(x) = 26.2 - 0.1x
f(100) = 26.2 - 0.1(100)
= 26.2 = 10.0
= 16.2
f(5-) = 26.2 - 0.1(50)
= 26.2 - 5.0
= 21.2
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Answer: D
<u>Step-by-step explanation:</u>
f(x) = 2|x - 3| + 1
2 units right: f(x) = 2|x - (3 + 2)| + 1
= 2|x - 5| + 1
Answer:
(8,6)
Step-by-step explanation:
count manually to get (8,6)
