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3 years ago

A spinning coil of wire is what moves the cone in a speaker, producing the sound. True/False

Physics

2 answers:

valentina_108 [34]3 years ago

6 0

Answer:

False

Explanation:

Georgia [21]3 years ago

5 0

False, the spinning coil of wire that moves the cone in a speaker does not produces sound.

<u>Explanation</u>:

The wire coil is an electromagnet that is fixed to speaker cone. A normal magnet attached to the back of the speaker cone.When audio is sent in the form of short bursts of electric current to the speaker cone through the wire.

A magnetic field is induced when electric current allowed to pass through the coil. This magnetic field is repelled by the other magnet. This repulsion cause the cone to move forward. In the absence of electric current in the coil, the cone moves backward.

Thus sound waves are produced due to the resulting rarefaction and compression. So it is not the spinning coil of wire but he permanent magnet that produces the sound.

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Four mass–spring systems oscillate in simple harmonic motion. Rank the periods of oscillation for the mass–spring systems from l

liubo4ka [24]

The periods of oscillation for the mass–spring systems from largest to smallest is:

m = 4 kg , k = 2 N/m (T = 8.89 s)
m = 2 kg , k = 2 N/m (T = 6.28 s)
m = 2 kg , k = 4 N/m (T = 4.44 s)
m = 1 kg , k = 4 N/m (T = 3.14 s)

<h3>Explanation:</h3>

The period of oscillation in a simple harmonic motion is defined as the following formulation:

$T=2\pi \sqrt{\frac{m}{k} }$

Where:

T = period of oscillation

m = inertia mass of the oscillating body

k = spring constant

m = 2 kg , k = 2 N/m

$T=2\pi \sqrt{\frac{2}{2} }$

$T=2\pi$

T = 6.28 s

m = 2 kg , k = 4 N/m

$T=2\pi \sqrt{\frac{2}{4} }$

$T=2\pi \sqrt{\frac{1}{2} }$

T = 4.44 s

m = 4 kg , k = 2 N/m

$T=2\pi \sqrt{\frac{4}{2} }$

$T=2\pi \sqrt{2 }$

T = 8.89 s

m = 1 kg , k = 4 N/m

$T=2\pi \sqrt{\frac{1}{4} }$

$T=\pi$

T = 3.14 s

Therefore the rank the periods of oscillation for the mass–spring systems from largest to smallest is:

m = 4 kg , k = 2 N/m (T = 8.89 s)
m = 2 kg , k = 2 N/m (T = 6.28 s)
m = 2 kg , k = 4 N/m (T = 4.44 s)
m = 1 kg , k = 4 N/m (T = 3.14 s)

Learn more about simple harmonic motion brainly.com/question/13058166

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4 0

3 years ago

Read 2 more answers

Two identical small charged spheres are a certain distance apart, and each initially experiences an electrostatic force of magni

Rom4ik [11]

Answer:

New force, $F'=\dfrac{1}{4}F$

Explanation:

The electrostatic force between two spheres is given by :

$F=\dfrac{kq_1q_2}{r^2}$

According to given condition, each of the spheres has lost half its initial charge, new force is given by :

$F'=\dfrac{kq_1/2q_2/2}{r^2}$

$F'=\dfrac{1}{4}\times \dfrac{kq_1q_2}{r^2}$

$F'=\dfrac{1}{4}F$

So, the new force becomes (1/4)th of the initial force. Hence, the correct option is (d).

4 0

3 years ago

Consider two spaceships, each traveling at 0.50c in a straight line. Ship A is moving directly away from the Sun and ship B is a

attashe74 [19]

Answer:

The velocity of the light will be 1.0c only

Explanation:

The velocity of the light measured in the case given in question will be 1.0c only.

This is due to the fact that the velocity of light is never relative. The velocity of the light is maximum

The velocity of the light cannot be scaled down in no case

Thus, the velocity of the light remains as constant.

Hence, the velocity of the light measured will be 1.0c although the ships have relative velocity.

3 0

2 years ago

Two thermometers are calibrated, one in de

zlopas [31]

-40 c = -40 f but k would be 233.15

6 0

3 years ago

Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding. Disregarding friction, what is the velo

lilavasa [31]

Answer:

The velocity of the Mr. miles is 17.14 m/s.

Explanation:

It is given that,

Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding, h = 15 m

We need to find the velocity of the Mr. Miles at the bottom of the slide. It is a case of conservation of energy which states that the total energy of the system remains conserved. Let v is the velocity of the Mr. miles. So,

$v=\sqrt{2gh}$

g is the acceleration due to gravity

$v=\sqrt{2\times 9.8\times 15}$

v = 17.14 m/s

So, the velocity of the Mr. miles is 17.14 m/s. Hence, this is the required solution.

5 0

3 years ago