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Anna007 [38]
2 years ago
5

As you know, loudspeakers are used for communication at sporting events, and in schools or supermarkets. Research loudspeakers o

n the Web. Describe the components of a speaker and explain how it produces sound. In particular, explain how the force on a current-carrying wire in a magnetic field is used to make a speaker operate.
Physics
1 answer:
Ivahew [28]2 years ago
4 0

Sound is invisible, but sometimes we can feel it. when a kettle-drum is thumped with a stick that time the tight drum skin moving up and down very quickly for some time and then it pumped sound waves into the air. Loudspeakers also work in a similar manner.

components of a speaker:

Electromagnet

In order to translate an electrical signal into an audible sound, an electromagnet is used in looudspeakers

 Suspension

suspension is used to  center the voice coil in the gap of the magnet and it exerts a restoring force to keep it there. suspension is used to limits the maximum mechanical excursion of the diaphragm and voice coil.

voice Coil

Voice coil is a coil of wire which is attached to the apex of a loudspeaker cone. It provides the motive force to the cone by the reaction of a magnetic field to the current passing through it.

Cone

In a  loudspeaker, a  thin, semi-rigid membrane attached to the voice coil, which moves in a magnetic gap due to which it vibrates  and produced sound. It is known as diaphragm and It can also be called a cone, though not all speaker diaphragms are cone-shaped.

Surround

surround is also known as front suspension. surround is used to join the cone to the chassis. Together with the suspension it controls the cone excursion, and it is also used to  determine how energy which is  travelling through the cone is absorbed, and how the speaker limits when it reaches the ends of its travel.

Dustcap

dustcap is used to protect the voice coil from dust and dirt. It is a part of the cone.

How speaker work

The outer part of the cone is fastened to the outer part of the loudspeaker's circular metal rim. The inner part is fixed to  voice coil that sits just in front of a permanent magnet . When loudspeaker is hooked up to a stereo, electrical signals feed through the speaker cables into the coil. It will convert the coil into an electromagnet magnet . when electricity flows back and forth in the cables, the electromagnet either repels or attract the permanent magnet. This moves the coil back and forth, and it pull and push the loudspeaker cone. Like a drum skin vibrating back and forth, the moving cone pumps sounds out into the air.In this way loudspeakers works.

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Lady_Fox [76]

Answer:

The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.

Explanation:

Knowing that the density of lead is 11,3 g/cm^{3} and the volume, we can calculate the true weight of the piece of lead:

weight_{lead}=\rho _{lead}*V_{lead}=11,3  g/cm^{3} *50 cm^{3}   = 565 g

When the experiment is done in air, we can discard buoyancy force (due to different densities) made by air because it's negligible and the measured weight is approximately the same as the true weight.

When it is done in water, the effect of buoyancy force (force made by the displaced water) is no longer negligible, so we have to take it into account.

Knowing that the density of water is 1 g per cubic centimeter, and that the volume displaced is equal to the piece of lead (because of its much higher density, the piece of lead sinks), we can know that the buoyancy force made by water is 50 g, opposite to the weight of the lead.

Weight_{measured}=weight_{lead}-weight_{water}=\frac{(565 g *9.8 m/s^{2}  -50 g*9.8 m/s^{2})}{9,8m/s^{2} }  = 515 g

Now that we have the two measurements, we can calculate the difference:

Difference= |Weight _{in   water}- Weight _{in   air}|=|515 g-565 g|=50 g

The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.

4 0
2 years ago
What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate
Alex_Xolod [135]

Answer:

Mass will be 4.437 kg

Explanation:

We have given force constant k = 7 N/m

Time period of oscillation T = 5 sec

So angular frequency \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec

We know that angular frequency is given by

\omega =\sqrt{\frac{k}{m}}

1.256 =\sqrt{\frac{7}{m}}

Squaring both side

1.577 =\frac{7}{m}

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2 years ago
In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg
harkovskaia [24]

Answer:

λ = 2042 nm

Explanation:

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solution

we will find first angle between first max and central bright

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tan θ = 4.5 ×10^{-2}  / 11

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L sinθ  = mλ

here we know m = 1  so put all value and find λ

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F = kq1q2/r^2

where:
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k = Coulomb's constant, 9x10^9 Nm^2/C^2
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r = distance between charges, meters

Using direct substitution, the force F is determined to be 1920 Newtons.</span>
7 0
2 years ago
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