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kow [346]
3 years ago
12

A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be in a s

peaker where the desired diffraction angle is 75° and a 9100 Hz sound is generated? The speed of sound is 343 m/s
Physics
1 answer:
Greeley [361]3 years ago
6 0

Answer:

a = 4.76 cm

Explanation:

As we know by the law of diffraction through circular aperture

a sin\theta = 1.22 \lambda

here we know that

a = diameter of the aperture

\theta = diffraction angle

\lambda = wavelength

now we have

a = \frac{1.22 \lambda}{sin\theta}

Now in order to find the wavelength we know

\lambda = \frac{c}{f}

\lambda = \frac{343}{9100}

\lambda = 0.0376 m

now we have

a = \frac{1.22(0.0376)}{sin75}

a = \frac{0.046}{0.966}

a = 0.0476 m= 4.76 cm

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kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:  

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t  Equation 2

y: The vertical distance the ball moves at time t  

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t  

Known information

We know the following data:

Vi=15 m / s

g =9.8 \frac{m}{s^{2} }

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)  

a) t=1s

y = 15*1 - ½ 9.8*1^{2}=15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

y = 15*1.5 - ½ 9.8*1.5^{2}=22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

y = 15*2 - ½ 9.8*2^{2}= 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

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Answer:

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Explanation:

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Two waves meet and overlap. The first wave has an amplitude of 4.6 centimeters and the second wave has an amplitude of 2 centime
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Answer:

Explanation:

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