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nekit [7.7K]
2 years ago
3

two loudspeakers facing each other are separated by 1.00 m and are excited in phase by a 1,100 Hz sine wave. Where will be the p

oints on a line between speakers where there will be constructive interference, and where will be the points where there will be destructive interference
Physics
1 answer:
faust18 [17]2 years ago
5 0

Answer:

Condition of constructive interference is given as

x = 95 cm, 80 cm, 65 cm, 50 cm, 35 cm, 20 cm, 5 cm

condition of destructive interference we know that

x = 87.5 cm, 72.5 cm, 57.5 cm, 42.5 cm, 27.5 cm, 12.5 cm

Explanation:

As we know that the frequency of the sound waves is 1100 Hz

now we can find the wavelength as

\lambda = \frac{c}{f}

now we have

\lambda = \frac{330}{1100}

\lambda = 30 cm

now we know for the condition of constructive interference

Path difference = N\lambda

(1 - x) - x = N(0.30)

1 - 2x = N(0.30)

x = \frac{1 - 0.30 N}{2}

x = 95 cm, 80 cm, 65 cm, 50 cm, 35 cm, 20 cm, 5 cm

Similarly for condition of destructive interference we know that

Path difference = (2N + 1)\frac{\lambda}{2}

so we have

(1 - x) - x = (2N + 1)(0.15)

1 - 2x = N(0.30) + 0.15

x = \frac{0.85 - 0.30 N}{2}

x = 87.5 cm, 72.5 cm, 57.5 cm, 42.5 cm, 27.5 cm, 12.5 cm

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Which of the following diagrams involves a virtual image ?
sergiy2304 [10]

Answer:

The third diagram

Explanation:

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The magnetic field around the head has been measured to be approximately 3.00×10−8 gauss . Although the currents that cause this
konstantin123 [22]

Answer:

3.81972\times 10^{-7}\ A

Explanation:

B = Magnetic field = 3\times 10^{-8}\ G

d = Diameter of loop = 16 cm

r = Radius = \frac{d}{2}=\frac{16}{2}=8\ cm

i = Current

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

The magnetic field of a loop is given by

B=\frac{\mu_0i}{2r}\\\Rightarrow i=\frac{B2r}{\mu_0}\\\Rightarrow i=\frac{3\times 10^{-8}\times 10^{-4}\times 2\times 0.08}{4\pi\times 10^{-7}}\\\Rightarrow i=3.81972\times 10^{-7}\ A

The current needed to produce such a field at the center of the loop is 3.81972\times 10^{-7}\ A

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Answer:

0.0367

Explanation:

The loss in kinetic energy results into work done by friction.

Since kinetic energy is given by

KE=0.5mv^{2}

Work done by friction is given as

W= umgd

Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.

Making u the subject of the formula then we deduce that

u=\frac {v^{2}}{2gd}

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367

Therefore, the coefficient of kinetic friction is approximately 0.0367

7 0
2 years ago
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