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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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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
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Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

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According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

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For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

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