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3 years ago

To what does the amplitude of a sound refer to?

Physics

1 answer:

Alex777 [14]3 years ago

7 0

Answer:

The number of molecules displaced by a vibration creates the amplitude of a sound. The strength or level of sound pressure. The number of molecules in the sound wave in b is greater than the number in the sound wave in a, therefore the amplitude of the sound wave in b is greater.

Explanation:

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A 2kg bowling ball is 2.5 meter off the ground on a post when it falls. Just before it reaches the ground, it is traveling 7m/s.

shutvik [7]

<span>The mechanical energy is conserved.

I hope this helps, good luck! :)</span>

7 0

4 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

4 years ago

The mass of a star is 1.210×1031 kg and it performs one rotation in 20.30 days. Find its new period (in days) if the diameter su

balandron [24]

The new period will be 2.486 days.

<h3>What is the period?</h3>

The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.

Given data;

Mass of a star,m= 1.210×10³¹ kg

The time period for one rotation of the star, T = 20.30 days

D' = 0.350 D

R' = 0.350 R

From the law of conservation of angular momentum;

$\rm I \omega = I' \omega' \\\\ \frac{2}{5} MR^2 \times \frac{2 \pi }{T}=\frac{2}{5} MR'^2 \\\\ R^2 \times \frac{1}{T}= R'^2 \times \frac{1}{T} \\\\ T' = \frac{R'^2T}{R^2} \\\\ T' = \frac{(0.350 R)^2 \times 26.1 }{R^2} \\\\T' = 0.1225 \times 20.30 \\\\ T'= 2.486 \ days$

Hence, the new period will be 2.486 days.

To learn more about the period, refer to the link;

brainly.com/question/569003

#SPJ1

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2 years ago

An iron has a power of 1,000w. How much energy will it electrically transfer each second?.

Mariana [72]

Answer:

P= 1000W, t = 1s ⇒ E = Pt = 1000 ∗ 1 = 1000J

Explanation: Power is the rate of doing work or of transferring heat per unit of time, or, in other words, power is the rate of transferred energy per unit of time

Explanation:

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1 year ago

Which of these does NOT emit visible light?

s344n2d4d5 [400]

c the galaxy because on a nice night you cant see the milkly way

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3 years ago