Does air play a role in the propagation of the human voice from one end of a lecture hall to the other

Physics

1 answer:

love history [14]3 years ago

4 0

Answer: Yes

Air serves as a propagation medium.

Explanation:

Propagation of sound wave require a medium for transmission. Sound is produced by vibrating objects, The matter or substance through which sound is transmitted is called a medium. It can be solid, liquid or gas. Therefore to propagate sounds (human voice) from one end of a lecture hall to the other we need a propagation medium which is air. Air serves as a propagation medium.

You might be interested in

How can electric energy be converted to heat energy?

Goshia [24]

To do that, you must pass electric current through a substance
that electrons have to spend energy to pass through.
The substance will be one that gets warm and dissipates heat
when electric current flows through it.
We'll say that the substance has "resistance", which we can measure.
The amount of heat that appears when current flows through it
will be (current²)·(resistance).

A few examples of things used for that purpose:

-- resistors
-- burners on electric stoves
-- coils of resistor-wire in a toaster
-- aquarium heater
-- electric clothes iron
-- electric coffee pot
-- blow-dryer
-- electric hair-curling iron
-- skinny tungsten wire in a light-bulb .

5 0

3 years ago

WOULUJUTUL RECIPECUIUS.

3241004551 [841]

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider $M_{1}$ and $M_{2}$ as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

$\text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}$

As gravitational constant $G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}$ , $M_{1}$ = 20 kg and $M_{2}$ = 100 kg, while d = 2.6 m, then