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 last choice is right sure because this is a perfect square trinomial
(4xy -3z)^2 = 16x^2y^2 -24xyz +9z^2
hope this will help you
Answer:
Step-by-step explanation:
2(5 + y) = 18
10 + y = 18
Bringing like terms on one side
Y = 18 - 10 = 8
Answer:
they are 4 i think
Step-by-step explanation:
5 x 4 = 20
Answer:
angle 3 = 79 degree
Step-by-step explanation:
given:
angle 1 = 98 degree
angle 2 : 19 degree
angle 3 =?
angle 2 + angle 3 =angle 1 (sum of two interior angle opposite angles is equal to the exterior angle formed)
19 + angle 3 = 98
angle 3 = 98 - 19
angle 3 = 79 degree
