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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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If you were in charge of designing a wire to carry electricity across your city, state or province, which of
Anon25 [30]

Answer:

Thin, aluminium and buried underground.

Explanation:

When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.

i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.

ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.

iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.

7 0
2 years ago
You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you ho
timurjin [86]

Answer:

\omega'=19.419\ rev.s^{-1}

Explanation:

Given:

angular speed of rotation of friction-less platform, \omega=5.1\ rev.s^{-1}

moment of inertia with extended weight, I=9.9\ kg.m^2

moment of inertia with contracted weight, I'=2.6\ kg.m^2

<u>Now we use the law of conservation of angular momentum:</u>

I.\omega=I'.\omega'

9.9\times 5.1=2.6\times \omega'

\omega'=19.419\ rev.s^{-1}

The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.

5 0
3 years ago
A piece of wood has mass of 250g and volume of 400cm cubed. what is its density​
Zigmanuir [339]

Answer:

0.625 \: g {cm}^{ - 3}

Explanation:

density \:  =  \frac{mass}{volume} \\  \\  =  \frac{250}{400}   \\  \\  = 0.625 \: g {cm}^{ - 3}

5 0
2 years ago
The gravitational force of attraction be-
Sliva [168]

Answer:

2.15 m

Explanation:

Newton's Law of Universal Gravitation:

\displaystyle F_g = G \frac{m_1 m_2}{r^2}  

  • \displaystyle F_g is the gravitational force of attraction
  • \displaystyle G is the universal gravitational constant
  • m_1 and m_2 are the two masses of the two objects
  • \displaystyle r is the distance between the centers of the two objects.

List the known values:

  • \displaystyle F_g = 3.47 \cdot 10^-^8 \ \text{N}  
  • \displaystyle G = 6.673 \cdot 10^-^1^1 \ \frac{Nm^2}{kg^2}
  • \displaystyle m_1 = 44.8 \ \text{kg} \\ m_2 = 53.9 \ \text{kg}
  • \displaystyle r =\ ?

Plug these values into the equation:

  • \displaystyle 3.47 \cdot 10^-^8 \ \text{N} = 6.673 \cdot 10^-^1^1 \ \frac{Nm^2}{kg^2} \frac{(44.8 \ \text{kg})(53.9 \ \text{kg})}{r^2}

Notice that the units \displaystyle \text{N}, \displaystyle \text{kg}^2, and \displaystyle \text{m} cancel out. We are left with the unit \displaystyle \text{m} for radius r.

Get rid of the units to make the problem easier to read.

  • \displaystyle 3.47 \cdot 10^-^8= 6.673 \cdot 10^-^1^1 \ \frac{(44.8)(53.9) \ }{r^2}  

Multiply the masses together.

  • \displaystyle 3.47 \cdot 10^-^8= 6.673 \cdot 10^-^1^1 \ \frac{(2414.72)}{r^2}

Multiply the gravitational constant and the masses together.

  • \displaystyle 3.47 \cdot 10^-^8= \frac{1.61134265\cdot 10^-^7}{r^2}

Solve for r^2 by dividing both sides by 3.47 * 10^(-8) and moving r^2 to the left.

  • \displaystyle r^2 = \frac{1.61134265\cdot 10^-^7}{3.47\cdot 10^-^8}
  • r^2 = 4.64363876

Take the square root of both sides.

  • r=2.154910383  

The students are sitting about 2.15 m apart from each other.

4 0
2 years ago
Question 2 - If Juan starts out at 10m/s, and in 15 s speeds up to 60 m/s, what is his acceleration?
34kurt

Answer:

B) 3.33 m/s²

Explanation:

Given that,

The initial velocity of Juan, u = 10 m/s

Final velocity of Juan,  v = 60 m/s

Time taken, t = 15 s

The acceleration is given by the relation

                          a = v-u/t  m/s²

Substituting in the above equation

                           a = 60 - 10 / 15

                           a = 3.33 m/s²

Hence, the acceleration of the Juan is, a = 3.33 m/s²

5 0
3 years ago
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