Answer:
Orange and pink, I'm not too sure
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:
Step-by-step explanation:
Ok let start by putting a set amount of hats, let start with 12, 8 times 12 equals 96, 200-96 is 104… I would subtract 8 until you get a number you can divide equally by 10, 104-8=96, 96-8 =88 , 88-8=80, there is 80 dollars left! Ok 80 divide by 10 = 8 so
8 shirts and we’ll we started at 12 and then we add how many more time we subtracted 8 so 15 total,
So, in total 8 shirts and 15 hats!!!
<span>3425 x 10^-2 =
34.25
maybe</span>
Answer:
51.2545454545
Step-by-step explanation:
