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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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a waterwheel built in Hamah, Syria, has a radius of 20 m . If the tangential velocity at the wheels edge is 7.85 m/s , what is t
gtnhenbr [62]

Answer:

3.08m/s²

Explanation:

Given parameters:

Radius = 20m

Tangential velocity  = 7.85m/s

Unknown:

Centripetal acceleration  = ?

Solution:

Centripetal acceleration is the acceleration of a body along a circular path.

 it is mathematically given as;

       a  = \frac{v^{2} }{r}  

v is the tangential velocity

r is the radius

      a  = \frac{7.85^{2} }{20}   = 3.08m/s²

4 0
2 years ago
When we see a meteor shower, it means that ________.
Irina18 [472]

Answer:

option A

Explanation:

The meteor shower is the celestial activity in which meteors are observed to radiate or originate from one point.

Meteors are nothing but dust or ice from the trails of comets. Most of the meteors are less than the size of the sand particle.

We will see comet shower when we earth will cross the orbit of the comet.

Hence, the correct answer is option A

6 0
3 years ago
Two parallel wires are separated by 6.10 cm, each carrying 2.85 A of current in the same direction. (a) What is the magnitude of
Westkost [7]

Answer:

The force per unit length is 2.66 \times 10^{-5} \ N/m

Explanation:

The current carrying by each wires = 2.85 A

The current in both wires flows in same direction.

The gap between the wires = 6.10 cm

Now we will use the below expression for the force per unit length. Moreover, before using the below formula we have to change the unit centimetre into meter. So, we just divide the centimetre with 100.

F/l = \frac{\mu _0i_1 i_2}{2\pi d} \\i_1 = 2.85 \\i_ 2 = 2.85  \\\mu _0 = 4\pi \times 10^{-7} \\d = 0.061 \\F/l = \frac{4\pi \times 10^{-7} \times 2.85 \times 2.85}{2 \pi \times 0.061} \\= 2.66 \times 10^{-5} \ N/m

4 0
3 years ago
Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?
Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

h = \frac{v^{2}}{2g}

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

\frac{mv^{2}}{2} = mgh

h = \frac{v^{2}}{2g}

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0
2 years ago
Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.
Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge q_{1}= 3.8\times10^{-6}\ C

Second charge q_{2}=3.2\times10^{-6}\ C

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

E_{1}=E_{2}

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}

\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}

\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}

x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}

Put the value into the formula

x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}

x=1.69\ m

Hence, The distance is 1.69 m.

5 0
3 years ago
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