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3 years ago

Characteristics of Sound Wave.

Physics

1 answer:

jeka943 years ago

4 0

Wavelength,
Amplitude
Time-Period,
Frequency
Velocity or Speed

Sound wave can be described by five characteristics: Wavelength, Amplitude, Time-Period, Frequency and Velocity or Speed. The minimum distance in which a sound wave repeats itself is called its wavelength. That is it is the length of one complete wave

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The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o

N76 [4]

Frequency = (speed) / (wavelength)

Frequency = (331 m/s) / (0.6 m) = 551.7 Hz

3 0

3 years ago

Help me answer this question and get yourself some points!

Marrrta [24]

Answer:

A×B=C×D

500×0.5=250×X

250=250×X

X=250/250=1

X=1 m

Explanation:

note: if the force plus two, the distance will be half.

4 0

3 years ago

A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago

Solve the problem.

gulaghasi [49]

As the shock waves travel in concentric outward circles from the epicenter, and the diameter is measured 120 miles,
area of a circle =πr*r

d=120
r=120/2r=6060*60=36003600*π=11309.734
11309.734 square miles

5 0

3 years ago

Car 1 brakes to a stop on a dry road. Car 2 does the same thing, at the same speed, on a road where it has been raining. Explain

Zepler [3.9K]

Answer: car 1 is going how fast

Explanation: no need to answer without speed I won't know distance.

8 0

3 years ago

Characteristics of Sound Wave.​

Characteristics of Sound Wave.