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:
No they did not pay the same rate, Bill paid more than Genevieve.
Step-by-step explanation:
Genevieve : 15 ÷ 4 =3.75
Genevieve paid 3.75 per ticket
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Bill : 75 ÷ 16 = 4.6875
Bill paid 4.6875 per ticket
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The future value of the investment in dollars s
V = p + prt
where
p = principal, dollars
r = annual interest rate (in decimal form)
t = time, years
To determine p, write the formula as follows:
Factorize p out on the right side.
V = p(1 + rt)
Divide each side b (1 + rt).
Answer:
