All categories

3 years ago

The displacement of a wave from the baseline.this affects thrloudness of sound

Physics

1 answer:

qwelly [4]3 years ago

5 0

Answer:

Amplitude is the fluctuation or displacement of a wave from its mean value. With sound waves, it is the extent to which air particles are displaced, and this amplitude of sound or sound amplitude is experienced as the loudness of sound. ... It means that maximum amount the wave varies from the baseline or equilibrium.

Explanation:

You might be interested in

14. Convert 22 degrees celsius to fahrenheit.<br> Please show your work

forsale [732]

(22°C × 9/5) + 32 = 71.6°F

6 0

3 years ago

Read 2 more answers

A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4

irina [24]

Here We can use principle of angular momentum conservation

Here as we know boy + projected mass system has no external torque

Since there is no torque so we can say the angular momentum is conserved

$mvL = (I + mL^2)\omega$

now we know that

m = 2 kg

v = 2.5 m/s

L = 0.35 m

I = 4.5 kg-m^2

now plug in all values in above equation

$2\times 2.5 \times 0.35 = (4.5 + (2\times 0.35^2))\omega$

$1.75 = [4.5 + 0.245]\omega$

$1.75 = 4.745\omega$

$\omega = 0.37 rad/s$

so the final angular speed will be 0.37 rad/s

4 0

3 years ago

PLEASE HELP!!!! :D When you look at the Sun through a filtered telescope, the visible portion of the Sun appears blotchy.

olya-2409 [2.1K]

<span>When an individual looks through a filtered telescope in which he or she observes the sun, the portion where it appears blotchy is likely to be called the sunspots while the layer of the sun where it shows where it occurs is called the photosphere.</span>

5 0

3 years ago

Read 2 more answers

As an object sinks in a fluid, the buoyant force ____.

kiruha [24]

Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.

As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.

3 0

3 years ago

Read 2 more answers

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the

Airida [17]

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

$V =\frac{Kq}{r}$

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V$

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V$

Total potential due to this charges = 4500 V + 6750 V = 11250 V

6 0

4 years ago