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:
- 4 + 8m
Step-by-step explanation:
4m - 4 + 4m = ?
4m + 4m = 8m
- 4 + 8m is the answer because you can not simplify it anymore.
Step-by-step explanation:
Tan30=4/adj
adj Tan 30= 4
adj=4/tan30
adj=6.93m
Area=L×b
4×6.93=27.72m
Answer:
I think the answers your look for are
A. 2 square root of 20
B. 6^1/2-5^1/2
C. 4*5^1/2
Step-by-step explanation: