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zlopas [31]
4 years ago
12

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.

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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

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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
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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

      So

         0 = 16.57 + 2 * a * 1.8

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

            a =  - 76.27 m/s^2

         

         

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3 years ago
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