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Tomtit [17]
4 years ago
7

A naval engineer is testing an nonreflective coating for submarines that would help them avoid detection by producing destructiv

e interference for sonar waves that have a frequency of 512 Hz and travel at 1390 m/s. If there is a phase change for waves reflected from both the top and bottom surfaces of the coating, what is the minimum thickness of the coating? For sonar waves let nwater=1.00 and ncoating=13.5. Express your answer to three significant figures and include appropriate units.
Physics
1 answer:
borishaifa [10]4 years ago
3 0

Answer:

5 cm .

Explanation:

Wave length of sonar waves λ = 1390 / 512

= 2.715 m

For waves getting reflected from both layers of the coating ,  undergoing phase reversal , net change will be zero .

For destructive interference

path diff = ( 2n+1) λ / 2

2μt =  ( 2n+1) λ / 2

2 x 13.5 x t =  ( 2n+1) λ / 2

for minimum thickness

n = 0

2 x 13.5 x t = λ / 2

2 x 13.5 x t =  λ / 2

t = λ / (2 x 27)

= 2.715 / 2 x 27

= .05 m

5 cm .

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Answer:

The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

Explanation:

Given that,

Angular acceleration \alpha=\pi\ rad/s^2

Angular displacement \theta=\pi\ rad

Angular velocity \omega =2\pi\ rad/s

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}

\omega_{0}^2=\omega^2-2\alpha\theta

Where, \alpha = angular acceleration

\omega = angular velocity

Put the value into the formula

\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi

\omega=\sqrt{2\pi^2}

\omega_{0}=\pi\sqrt{2}\ rad/s

Hence, The angular velocity at the beginning of the interval is \pi\sqrt{2}\ rad/s.

3 0
3 years ago
Find the frequency of a wave with the wavelength 3.5 m and the speed is 50 m/s. <br>​
Andru [333]

Answer:

8.57 Hz

Explanation:

From the question given above, the following data were obtained:

Wavelength (λ) = 3.5 m

Velocity (v) = 30 m/s

Frequency (f) =?

The velocity, wavelength and frequency of a wave are related according to the equation:

Velocity = wavelength × frequency

v = λ × f

With the above formula, we can simply obtain the frequency of the wave as follow:

Wavelength (λ) = 3.5 m

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Frequency (f) =?

v = λ × f

30 = 3.5 × f

Divide both side by 3.5

f = 30 / 3.5

f = 8.57 Hz

Thus, the frequency of the wave is 8.57 Hz

7 0
3 years ago
Space-shuttle astronauts experience accelerations of about 35 m/s2 during takeoff. What force does a 75 kg astronaut experience
amm1812

Answer:

<h3>The answer is 2625 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 75 × 35

We have the final answer as

<h3>2625 N</h3>

Hope this helps you

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Answer:

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newton's second law of motion is represented using: f=ma

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from f=ma,

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