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Bogdan [553]
3 years ago
6

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

speaker where the desired diffraction angle is 11° and a 9100 Hz sound is generated? The speed of sound is 343 m/s.
Physics
1 answer:
andriy [413]3 years ago
6 0

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

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

Here,

\lambda = Wavelength

d= Diameter of aperture

\theta = Angular resolution or diffraction angle

Our values are given as,

\theta = 11\°

The frequency of the sound is f = 9100 Hz

The speed of the sound is v = 343 m/s

The wavelength of the sound is

\lambda = \frac{v}{f}

Here,

v = Velocity of the wave

f = Frequency

Replacing,

\lambda = \frac{(343 m/s)}{(9100 Hz)}

\lambda = 0.0377 m

The diffraction condition is then,

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

Replacing,

sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}

d = 0.24 m

Therefore the diameter should be 0.24m

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Answer:

K = 1800 kJ

Explanation:

Given that,

The speed of the object, v = 30 m/s

Mass of the object, m = 4000 kg

We need to find the kinetic energy of the object. The formula for the kinetic energy is given by :

K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 4000\times 30^2\\\\K=1800000\\\\or\\\\K=1800\ kJ

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A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin
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(1)

Cheetah speed: v_c = 97.8 km/h=27.2 m/s

Its position at time t is given by

S_c (t)= v_c t (1)

Gazelle speed: v_g = 78.2 km/h=21.7 m/s

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

S_g(t)=S_0 +v_g t (2)

The cheetah reaches the gazelle when S_c=S_g. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

v_c t=S_0 + v_g t

(v_c -v_g t)=S_0

t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s


(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

Cheetah: S_c = v_c t =(27.2 m/s)(7.5 s)=204 m

Gazelle: S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m

So, the gazelle should be ahead of the cheetah of at least

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