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:
D) 6
Step-by-step explanation:
set Q has 16 values which are all even values from 4 to 36
set Z has 11 values that are multiples of 3 from 4 to 36 (6,9,12,15,etc.) however, we have to eliminate the odd ones, leaving the following six:
6, 12, 18, 24, 30, 36
Answer:
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Answer:
C
Step-by-step explanation:
Answer:
When AB coincides with AC, the boundary of EAF maps exactly onto the boundary of CAD, implying that EAF dilates into CAD. So, the boundaries of EAF and CAD are similar.
The proportional relationship may be stated in different forms, but should be equivalent to this equation:
radius of Ab/ radius of Ac= length of arc EF/ length of arc CD
Explanations will vary, but should be based on the similarity of EAF and CAD. The proportional relationship follows from the fact that corresponding pairs of lengths in two similar figures have the same ratio.
Step-by-step explanation:
This is the exact answer so make sure you change it up a little.