Physics, 13.01.2020 03:31 zamawi17

Which best contrasts electric doorbells and audio speakers?

doorbells involve rapid changes in the direction of current, and speakers do not.
speakers involve rapid changes in the direction of current, and doorbells do not.
doorbells require closed circuits in order to work, and speakers do not.
speakers require closed circuits in order to work, and doorbells do not.

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Answer from: timozy95

The only true statement among the choices is:

Speakers involve rapid changes in the
direction of current, and doorbells do not.

The other three statements are false.

Answer from: leilaford2003

Answer;

Speakers involve rapid changes in the direction of current, and doorbells do not.

Explanation;Speakers are made of a permanent magnet surrounding an electromagnet which is attached to the loudspeaker membrane or cone. By varying the electric current through the wires around the electromagnet, the electromagnet and the speaker cone can move back and forth. If the variation of the electric current is at the same frequencies of sound waves, the resulting vibration of the speaker cone will create sound waves, including that from voice and music.Electric door bells use electromagnets, which is a temporary magnet whose magnetism can be varied by changing the amount of current. pressing the switch closes the contact and thus completing the circuit and the gong hits the bell breaking the circuit again. Both the speakers and the doorbells use the concept of electromagnetism however, they differ in that the speakers involve rapid changes in the direction of current, and doorbells do not, instead, it relies on completing and breaking the circuit.

Answer from: Quest

the gas collects in coils that then becomes a liquid that then flows through the coils inside the refrigerator

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .