How are loudness and intensity related to the amplitude and energy of a sound wave? What is the unit of intensity?

1 answer:

6 0

WattsIn the SI system, it has units watts per square metre (W/m2). It is used most frequently with waves (e.g. sound or light), in which case the average power transfer over one period of the wave is used. Intensity can be applied to other circumstances where energy is transferred.the greater the amplitude, the louder the sound, and the more energy there is.

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A planet has two

lozanna [386]

Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.

Moon #1: (1.262 days)² / (2.346 x 10^4 km)³

Moon #2: (orbital period)² / (9.378 x 10^3 km)³

If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.

Cross multiply the proportion:

(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³

Divide each side by (2.346 x 10^4)³:

(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³

= 0.1017 day²

Orbital period = 0.319 Earth day = about 7.6 hours.

7 0

3 years ago

Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity

Anna11 [10]

Answer:

the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

the distance that A slides relative to B is $S = \frac{v_o^2M}{2 \mu g (M+m)}$

Explanation:

From the diagram below;

acceleration of A relative to B is : $a = - ( \mu g + \frac{ \mu mg}{M})$

where

v = u + at

$0 = v_o + ( - \mu g - \frac{\mu m g }{M})t$

Making t the subject of the formula; we have:

$t = \frac{v_o M}{(\mu g )(M+m)}$

$v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\$

$S = \frac{v_o^2M}{2 \mu g (M+m)}$ which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

$v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\$

$v = \frac{mv_o}{m+M}$

Thus, the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

4 0

3 years ago

A ball is dropped from a height of 10m. At the same time, another ball is thrown

soldi70 [24.7K]

5.1 m

Explanation:

Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by

$y_1 = 10 - \frac{1}{2}gt^2$ (1)

where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground $y_2,$ is given by