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Tomtit [17]
3 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]3 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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<h3>The answer is 2.0 g/mL</h3>

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A 2-kg box is pushed to the right by a force of 4 N for a distance of 32 m. It has an initial velocity of 4 m/s to the right. NO
rewona [7]

Answer: a) 8 Kg m/s b) 16 Kg m/s c) 24 Kg m/s d) 16 J e) 128 J f) 144 J

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a) As momentum by definition is the product of mass times the velocity (is a vector quantity), we can write in this case the following:

pi = m. v₀ = 2 Kg . 4 m/s = 8 Kg. m/s

b) In order to get the change in momentum, we need to get first the final speed of the object.

As we have the total distance travelled, and we could find the acceleration, we could use a kinematic equation to solve the question, but later we will need the kinetic energy, it would be better to apply the work-energy theorem, and calculate ΔK as the work done by external force F, as follows:

ΔK = F . d = 1/2 m (vf² - v₀²)

As we know F, d, m, and v₀, we can solve the equation above for vf:

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pf = m . vf = 2 Kg. 12 m/s = 24 Kg. m/s

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e) We can find the change in the kinetic energy taking directly the difference between the final and initial ones, as follows:

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f) From above, we have Kf = 1/2 m. vf² = 1/2 . 2 Kg. 12² m²/s² = 144 J

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Fext = m .a ⇒ a = F/m = 4 N / 2 Kg = 2 m/s²

Appying the definition of acceleration, we can solve for t, as follows:

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