Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a be
1 answer:
Answer:
The frequency is
Explanation:
From the question we are told that
The frequency for the first note is
The beat frequency of the first note is
The frequency for the second note is
The beat frequency of the first note is
Generally beat frequency is mathematically represented as
Where are frequencies of two sound source
Now in the case of this question
For the first note
Where F is the frequency of the string note
For the second note
Adding equation 1 from 2
substituting values
=>
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Answer:
a) v_{p} = 2.83 m / s , b) 50.5º north east
Explanation:
This is a vector problem.
The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground
To make the sum we decompose the speed of the ball in its components
The angle of 30 east of the south, measured from the positive side of the x axis is
θ = 30 + 270 = 300
=
cos 300
= v_{b} sin. 300
v_{bx} = 3.60 cos 300 = 1.8 m / s
v_{by} = 3.60 sin 300 = -3,118 m / s
Let's add speeds on each axis
X axis
vₓ = v_{bx}
vₓ = 1.8 m / s
Y Axis
= v1 - vpy
v_{y} = 5.30 - 3.118
v_{y} = 2.182 m / s
The magnitude of the velocity can be found using the Pythagorean theorem
= √ (vₓ² + v_{y}²)
v_{p} = √ (1.8² + 2.182²)
v_{p} = 2,829 m / s
v_{p} = 2.83 m / s
b) for direction use trigonometry
tan θ = / vₓ
θ = tan ⁺¹ v_{y} / vₓ
θ = tan⁻¹ 2.182 / 1.8
Tea = 50.48º
This address is 50.5º north east
Answer:
The angular magnification is
Explanation:
From the question we are told
The focal length is
The near point is
The angular magnification is mathematically represented as
Substituting values