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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
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Answer: g= 25 oranges
Step-by-step explanation:
.38g + 15 = 24.5
.38g = 24.5 - 15
.38g = 9.5
g = 9.5 ÷ .38
g = 25
a. The formula solved for t is t = I/Pr
b. The value of t in the table is 3 years
<h3>Simple Interest</h3>
From the question, we are to solve for t in the given formula
The given formula is the formula for simple interest
I = Prt
To solve for t, we will divide both sides of the equation by Pr
That is,
I/Pr = Prt/Pr
I/Pr = t
∴ t = I/Pr
The formula solved for t is t = I/Pr
b. We are to find the value of t when
I = $75
P = $500
r = 5% = 0.05
From
t = I/Pr
t = 75/(500×0.05)
t = 75/25
t = 3 years
Hence, the value of t in the table is 3 years
Learn more on Simple interest here: brainly.com/question/25793394
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