All categories

4 years ago

In world war 1 , Multiple reflection was used in submarines. How multiple reflection works and helps in such situation?

Physics

1 answer:

prisoha [69]4 years ago

8 0

<u>Answer:</u>

Submarines use a device called Periscope that uses the concept of multiple reflections and help us see objects above the water surface.

<u>Explanation:</u>

When a light ray falls on a reflecting surface like a mirror, it gets reflected. In multiple reflections, the incident light ray is made to reflect multiple times by arranging the reflecting surfaces in different ways.

In submarines, we use Periscope, which is a long tube like structure. The long tube is bent at ends. It uses two simple mirrors which are placed parallel to each other at an angle of 45 degrees. The light from one mirror gets reflected to the other mirror, thus causing a multiple reflection.

You might be interested in

A bicycle is traveling north at 5.0 m/s. the mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radi

vova2212 [387]

Answer:

H(angular momentum) = I * ω inertia * angular speed

ω = V / R = 5 m/s / .2 m = 25 / s

I = M R^2 = 2 kg * (.2 m)^2 = .08 kg-m^2

H = 25 / s * .08 kg-m^2 = 2 kg-m^2/s

Using the right hand rule - the angular momentum is towards the west - the same direction as the angular velocity

7 0

2 years ago

saw5 [17]

U can infer that it’s a graph line

7 0

3 years ago

True or false

Kisachek [45]

Answer:

a) we should play inside our classroom

Explanation:

8 0

3 years ago

Read 2 more answers

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r

Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

$t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s$

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

$d=ut-\dfrac{1}{2}gt^2$

Putting all values

$d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m$

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0

3 years ago

A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming

Monica [59]

Answer:

The deceleration is $a = - 76.27 m/s^2$

Explanation:

From the question we are told that

The height above firefighter safety net is $H = 14 \ m$

The length by which the net is stretched is $s = 1.8 \ m$

From the law of energy conservation

$KE_T + PE_T = KE_B + PE_B$

Where $KE_T$ is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

and $PE_T$ is the potential energy of the before jumping which is mathematically represented at

$PE_T = mg H$

and $KE_B$ is the kinetic energy of the person just before landing on the safety net which is mathematically represented at

$KE_B = \frac{1}{2} m v^2$

and $PE_B$ is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

So the above equation becomes

$mgH = \frac{1}{2} m v^2$

=> $v = \sqrt{2 gH }$

substituting values

$v = 16.57 m/s$

Applying the equation o motion

$v_f = v + 2 a s$

Now the final velocity is zero because the person comes to rest

$0 = 16.57 + 2 * a * 1.8$

$a = - \frac{16.57^2 }{2 * 1.8}$

$a = - 76.27 m/s^2$

4 0

3 years ago