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3 years ago

The speed of sound is 330 m/s and the wavelength of a particular sound wave is 33 meters. Calculate the period of this wave.

Physics

2 answers:

weeeeeb [17]3 years ago

8 0

This one is easy it is 11

Anna11 [10]3 years ago

8 0

Answer:

The period of this wave es 10 seconds.

Explanation:

The wavelength is the minimum distance between two points of the wave that is in the same state of vibration.

Frequency is the number of vibrations that occur in a unit of time.

The speed of propagation is the speed with which the wave propagates in the middle. Relate wavelength (λ) and frequency (f) inversely proportionally using the following equation: v = f * λ.

The period is the time that elapses between the emission of two consecutive waves. In other words, it is the time it takes for a complete wave to pass through a reference point. The period (T) is simply the inverse of the frequency (f):

$T=\frac{1}{f}$

The units of the period are the seconds (s)

Then the propagation speed will be v = λ / T (Speed = wavelength/period)

Then T= λ / v

$Period = \frac{speed}{wavelength} = \frac{330 \frac{m}{s} }{33 m}$

Period (T)= 10 s

The period of this wave es 10 seconds.

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D. A fetus will develop in the fallopian tube.

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3 years ago

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A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

Umnica [9.8K]

Answer:

Explanation:

Given

Charge Q is uniformly spread over large non-conducting Elastic sheet

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$E=\frac{\sigma }{2\epsilon }$

where $\sigma =$ surface charge density $=\frac{q}{d^2}$

$E=\frac{\frac{q}{d^2}}{2\epsilon }$

for side 2d Electric Field is given by

$E'=\frac{\frac{q}{2d^2}}{2\epsilon }$

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3 years ago

A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can

belka [17]

Answer: 21.91 s

Explanation:

Given that,

Maximum height of the car, h = 48 ft

Acceleration of the elevator, a = 0.6 ft/s²

Deceleration of the elevator, -a = 0.3 ft/s²

Maximum speed of the elevator, v = 8 ft/s

Initial speed of the elevator, u = 0

If when the elevator accelerate from 0 to maximum velocity, v.

Let s be the vertical distance traveled during acceleration.

v² = u² - 2as

s = (v² - u²) / 2a

s = (8² - 0) / 2*0.6

s = 64 / 1.2

s = 53.33 ft

If when the elevator decelerates from maximum velocity, v to zero.

Let S be the vertical distance traveled during deceleration

u² = v² + 2aS

S = (u² - v²) / 2a

S = (0 - 8²) / 2 * 0.3

S = -64 / 0.6

S = 106.67 ft

Since he sum of s and S (i.e s + S) is greater than 48 ft, then the elevator will switch from acceleration to deceleration

without reaching the maximum velocity. Below, the switching point is labeled y.

v² = u² + 2ay

y = v²/2a

Inserting this into the earlier deceleration equation, we have

-v²/2 = d * [48 - (v²/2a)], where

d = deceleration

a = acceleration

Therefore, v = [4.√6. a √-(a.b/a)] / b

Where b = acceleration - deceleration

v = 4.382 ft/s

Using this newly found v, we proceed to find our s

s = (u² + v²)/2a

s = 19.2 / 1.2

s = 16 ft

The transport times for each segment are found from

v = u + a*t, thus upward t1

4.382 = 0 + 0.6 * t

t = 4.382/0.6

t = 7.303 s

Also,

4.382 = 0 + 0.3 * T

T = 4.382/0.3

T = 14.607 s

The total travel time is then t + T =

7.303 + 14.607

Total time of travel is 21.91 s

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3 years ago

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Ludmilka [50]

Answer: magnitude of the instantaneous angular velocity

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