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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Is there an image so I can see
P = kq/r
5 = k*1/2
k = 2*5 = 10
p = 5*10/50 = 1
P is 1.
Hope it helps!
Answer:
(f - g)(x) = -x² + 2x + 8
Step-by-step explanation:
Step 1: Plug in variables
2x + 1 - (x² - 7)
Step 2: Distribute the negative
2x + 1 - x² + 7
Step 3: Combine like terms
-x² + 2x + 8
