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STALIN [3.7K]
3 years ago
14

n noisy factory environments, it's possible to use a loudspeaker to cancel persistent low-frequency machine noise at the positio

n of one worker. The details of practical systems are complex, but we can present a simple example that gives you the idea. Suppose a machine 6.0 m away from a worker emits a persistent 90 Hz hum. Part A To cancel the sound at the worker's location with a speaker that exactly duplicates the machine's hum, how far from the worker should the speaker be placed
Physics
1 answer:
lys-0071 [83]3 years ago
4 0

Answer:

7.888m

Explanation:

Given:

frequency 'f'= 90Hz

velocity of the sound 'v' = 340 m /sec

The wavelength of the wave is given by,

λ= v/f => 340/90

λ= 3.777m

The destructive interference condition is givn by

Δd= (m+\frac{1}{2}  ) λ

where, m=0,1,2,3,..

m=0, for minimum destructive interference

Δd= (0+\frac{1}{2}  ) x 3.777

Δd=1.888

Therefore, the required distance is

d_f=d_i + Δd => 6 + 1.888

d_f= 7.888m

Thus, So the speaker should be placed at  7.888 m

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P before = P after = m₁ v₁ + m₂ v₂  

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4 years ago
A single conservative force F = (7.0x - 11) N, where x is in meters, acts on a particle moving along an x axis. The potential en
Umnica [9.8K]

Answer:

(a) 34.6429J

(b) -1.57 m

(c) 4.71 m

Explanation:

The derivative of the potential energy with respect to the position is equal to the negative of the force, so:

-\frac{dU}{dx}=F\\ dU=-Fdx

Then, if we integer both sides, we get:

∫dU = -∫(7x - 11)dx

U=\frac{-7}{2}x^{2} + 11x + c

we know that U is equal to 26 J when x is zero, so:

U=\frac{-7}{2}x^{2} + 11x + c

26=\frac{-7}{2}(0)^{2} + 11(0) + c

26=c

Finally, the equation for the potential energy is:

U(x)=\frac{-7}{2}x^{2} + 11x + 26

Therefore, the maximum positive potential energy is the energy when x is equal to 11/7. That is because the equation of U is the equation of a parable, and the vertex in a parable is given by:

x=\frac{-b}{2a} = \frac{-11}{2(-7/2)} =\frac{11}{7}

Where b is the number beside x and a is the number beside x^{2}, Then, the value of maximum U is:

U(11/7)=\frac{-7}{2}(11/7)^{2} + 11(11/7) + 26

U(11/7)=34.6429J

On the other hand, the negative and positive values of x where the potential energy is equal to zero is calculated as:

U(x)=\frac{-7}{2}x^{2} + 11x + 26

0=\frac{-7}{2}x^{2} + 11x + 26

if we solve this using the quadratic equation, we get:

x =\frac{-11+\sqrt{11^{2}-4(-7/2)(26)} }{2(-7/2)}=-1.5747

x =\frac{-11-\sqrt{11^{2}-4(-7/2)(26)} }{2(-7/2)}=4.7175

Finally, the negative and positive values of x where the potential energy is equal to zero are -1.5747 and 4.7175 respectively.

3 0
3 years ago
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