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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the answer to 54,000 x 10 is 540,000
You add all of them together if it’s labeled with a X.
For example, the first X is 1 hour. If it has 2 Xs it would be 2 hours, if no X is shown above, it’s zero hours.
Answer is 12 hours.
Answer:
AC = 25 units
Step-by-step explanation:
Given ∠ A = ∠ C then the triangle is isosceles and BC = AB , that is
7x - 1 = 5x + 5 ( subtract 5x from both sides )
2x - 1 = 5 ( add 1 to both sides )
2x = 6 ( divide both sides by 2 )
x = 3
Then
AC = 6x + 7 = 6(3) + 7 = 18 + 7 = 25 units
