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
#SPJ1
Answer:
5 feet
Step-by-step explanation:
Add 0.25 to both sides of the equation:
4.75 +0.25 = x - 0.25 +0.25
5.00 = x
The original height of the water in the pool was 5.00 feet.
Answer:
Area of room = 21 * 16
= 336 m²
Area of Carpet = (21 - 1)* (16-1)
= 20*15
= 300
Area of border = Area of room - Area of carpet
= 336 - 300
<h3>= 36 m².</h3>
Answer:
expanded form: 5 * 5 * 5
value: 125
Step-by-step explanation:
Answer:
I hope it will help you..
