Answer:
The possible approximate frequencies of the guitar string are 516.82 Hz and 508.82 Hz
Step-by-step explanation:
The given period of the tuning fork, T₁ = 1.95 × 10⁻³ s
The number of beats in 6.0 s = 24 beats
Therefore, the beat frequency, 
= 4 beats per second = 4 Hz
Let f₁ represent the frequency of the tuning fork, and let f₂ represent the frequency of the guitar string, we have;
f₁ = 1/T₁ = 1/(1.95 × 10⁻³) ≈ 512.82 Hz
We have;
Therefore, we get;
f₂ - 512.82 = ± 4
∴ f₂ = 512.82 ± 4
The possible frequencies of the guitar string, f₂ ≈ 516.82 Hz or f₂ ≈ 508.82 Hz
Answer:
.36666667
Step-by-step explanation:
Answer:
14
Step-by-step explanation:
you have opposite but need to find hypotenuse so you use SOH (sine=opposite/hypotenuse)
so sin(30)= 7/a
7/sin(30)=14
Hope its right
Answer:
yes it is parallel
Step-by-step explanation:
because I think I know it
Answer:
Explain how the Quotient of Powers was used to simplify this expression. (2 points) 25 = 22 By finding the quotient of the bases to be 1 4 and cancelling common factors F O
Step-by-step explanation:
By finding the quotient of the bases to be 1 4 and simplifying the expression O By simplifying 8 to 23 to make both powers base two, and subtracting the exponents O By simplifying 8 to 23 to make both powers base two, and adding the exponents 3.(MC)
