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babymother [125]

4 years ago

12

Create a mathematical model for the pressure variation as a function of position and time for a sound wave, given that the wavel

ength of the wave is λ = 0.190 m and the maximum pressure variation is ΔPmax = 0.270 N/m2. Assume the sound wave is sinusoidal. (Assume the speed of sound is 343 m/s. Use the following as necessary: x and t. Assume ΔP is in Pa and x and t are in m and s, respectively. Do not include units in your answer.)

1 answer:

goblinko [34]4 years ago

3 0

Answer:

The equation of position and time for a sound wave is $\Delta p=0.270(33.06 x-11342.40 t)$ .

Explanation:

Given that,

Wavelength = 0.190 m

Maximum pressure $\Delta P_{max}= 0.270 N/m^2$

We know that,

The function of position and time for a sound wave,

$\Delta p=\Delta p_{max}(kx-\omega t)$ ....(I)

We need to calculate the frequency

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{343}{0.190}$

$f=1805.2\ Hz$

We need to calculate the angular frequency

Using formula of angular frequency

$\omega =2\pi f$

Put the value into the formula

$\omega=2\pi\times1805.2$

$\omega=11342.40\ rad/s$

We need to calculate the wave number

Using formula of wave number

$k = \dfrac{2\pi}{\lambda}$

Put the value into the formula

$k=\dfrac{2\pi}{0.190}$

$k=33.06$

Now, put the value of k and ω in the equation (I)

$\Delta p=0.270(33.06 x-11342.40 t)$

Hence, The equation of position and time for a sound wave is $\Delta p=0.270(33.06 x-11342.40 t)$ .

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Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2.

IrinaVladis [17]

Answer:

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3 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

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$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (6)

Where:

$a=-4.9 m/s^{2}$

$b=48.1 m/s$

$c=1.5 m$

Substituting the known values:

$t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)}$ (7)

Solving (7) we find the positive result is:

$t=9.847 s$ (8)

Substituting this value in (1):

$x=(48.1 m/s)cos(0\°) (9.847 s)$ (9)

$x=473.640 m$ This is the horizontal distance the cannonball traveled before it landed on the ground.

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