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Nataliya [291]
3 years ago
5

A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.

Physics
1 answer:
fenix001 [56]3 years ago
5 0

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

(c) 352.9 Hz

Explanation:

For an open pipe,  the velocity is given by

v=\frac {2Lf}{n}

Making L the subject then

L=\frac {nV}{2f}

Where f is the frequency,  L is the length,  n is harmonic number,  v is velocity

Substituting 1 for n,  320 Hz for f and 331 m/s for v then

L=\frac {1*331}{2*320}=0.5171875\approx 0.52m

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, f2=2\times \frac {v}{2L} and f3=3\times \frac {v}{2L}

f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz

(c)

When v=367 m/s then

f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz

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A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
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Answer:

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This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

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