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:
Rule Example
+(+) Two like signs become a positive sign 3+(+2) = 3 + 2 = 5
−(−) 6−(−3) = 6 + 3 = 9
+(−) Two unlike signs become a negative sign 7+(−2) = 7 − 2 = 5
−(+) 8−(+2) = 8 − 2 = 6
In order to get the answer you will have to divide 30 by 100 then divide the number you get from that by 20 to receive your answer.
20/x=100/30
(20/x)*x=(100/30)*x
20=3.33333333333*x
(3.33333333333) to get x
20/3.33333333333=6
=6
Therefore your answer is "6."
Hope this helps
Answer:
50
Step-by-step explanation:
must use PEMDAS (order of operations)
step 1: multiply 7 and 6
8 + 42 / 3 x 3
step 2: divide 42 by 3
8 + 14 x 3
step 3: multiply 14 and 3
8 + 42
Add to get 50
Answer:D
Step-by-step explanation:
