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:
85.9 m
Step-by-step explanation:
The law of sines can help figure this.
The remaining angle in the triangle is ...
180° -75° -68° = 37°
This is the angle opposite the leg from the surveyor to the second marker. Referencing the attachment, we have ...
b/sin(B) = c/sin(C)
b = sin(B)·c/sin(C) = 132.3·sin(37°)/sin(68°) ≈ 85.873 . . . meters
The surveyor is about 85.9 meters from the second marker.
Answer:
4x+3y=68
9x+2y=77
plain= 5 shiny=16
Step-by-step explanation:
plain=x shiny=y
Product = multiply
3/4 x 12 = 9
17 + 9 = 26
