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3 years ago

What is the relationship between amplitude and volume?

Physics

1 answer:

MrRissso [65]3 years ago

5 0

Answer:

Explanation:

The sound moves in the form of waves. The amplitude is the distance between the highest and the lowest point of a wave. In this way the amplitude indicates the amount of energy that a sound signal contains.

Intensity is the amount of acoustic energy that a sound contains. Intensity is measured in decibels. Volume is a measure of the energy that a signal carries, being a magnitude of intensity.

In this way it is possible to say that the energy of a signal is closely related to its amplitude, but its development over time is also important.

The tone or height is the quality that distinguishes between a high or low sound and a low or high sound.

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An Olympic runner competing in a long-distance event finishes with a time of 2 hours, 45 minutes, and 35 seconds. The event has

Readme [11.4K]

Answer:

Average speed of Olympic runner will be 23.08 km/hr

Explanation:

We have given distance d = 27.3 miles

We know that 1 mile = 1.6 km

So 27.3 mile = $=27.3\times 1.6=43.68km$

Given time = 2 hour 45 minutes and 35 sec

We know that 1 hour = 60 minutes

So 45 minutes $=\frac{45}{60}=0.75hour$

We also know that 1 hour = 3600 sec

So 1 sec $=\frac{1}{3600}=2.777\times 10^{-4}hour$

So total time t = $=2+0.75+0.000277=2.750277hour$

We know that speed $=\frac{distance}{time }=\frac{63.48}{2.750277}=23.08km/hour$

6 0

4 years ago

A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and it

soldi70 [24.7K]

Answer:

The position function is $s_{t}=2t^3+5t^2-5t+9$ .

Explanation:

Given that,

Acceleration $a =12t+10$

Initial velocity $v_{0} = -5\ cm/s$

Initial displacement $s_{0}=9\ cm$

We know that,

The acceleration is the rate of change of velocity of the particle.

$a = \dfrac{dv}{dt}$

The velocity is the rate of change of position of the particle

$v=\dfrac{dx}{dt}$

We need to calculate the the position

The acceleration is

$a_{t} = 12t+10$

$\dfrac{dv}{dt} = 12t+10$

$a_{t}=dv=(12t+10)dt$

On integration both side

$\int{dv}=\int{(12t+10)}dt$

$v_{t}=6t^2+10t+C$

At t = 0

$v_{0}=0+0+C$

$C=-5$

Now, On integration again both side

$v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt$

$s_{t}=2t^{3}+5t^2-5t+C$

At t = 0

$s_{0}=0+0+0+C$

$C=9$

$s_{t}=2t^3+5t^2-5t+9$

Hence, The position function is $s_{t}=2t^3+5t^2-5t+9$ .

7 0

3 years ago

A 2130 kg car is parked on a hill that makes a 15º angle with the horizontal. What is the normal force on the car?

valentinak56 [21]

Answer:

20573.67N

Explanation:

Given;

mass (m) of the car = 2130kg

angle of inclination Θ = 15⁰

The normal force (F) on the car is given by

F = mgcosΘ

where g is the acceleration due to gravity.

Taking g as 10 $m/s^{2}$ and substituting the values of m and Θ into the equation. We have;

F = 2130 x 10 x cos 15⁰

F = 2130 x 10 x 0.9659

F = 20573.67N

Therefore the normal force on the car is 20573.67N

7 0

4 years ago

How did the projected storm track change from image A TO B TO C TO D

Gnom [1K]

Link provided in other answer is a SCAM, don’t click on the link !!!!

5 0

3 years ago

Find the unit vector in the direction of the resultant vector A=2i-3j+4k B=i+j+2k C=3i-2j+4k

shtirl [24]

Answer:

R = 6i - 4j + 10k

Explanation:

R = (2 + 1 + 3)i + (-3 + 1 - 2)j + (4 + 2 + 4)k

4 0

3 years ago

Read 2 more answers