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zlopas [31]
3 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]3 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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A 45.0 kg ice skater needs a 25 N horizontal force to get moving on a smooth ice surface. What is the coefficient of friction be
SVEN [57.7K]
The horizontal force : f = k*N
k- coefficient of friction
k = f /N
N = m * g = 45 kg * 9.81 m/s² = 441.45 N
k = 25 N : 441.45 N = 0.057
Answer C) 0.057
7 0
3 years ago
Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par
Murrr4er [49]

Answer:

X₃₁ = 0.58 m  and  X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

        F₁₃ - F₂₃ = 0

        F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

        F₁₃ = k q₁ q₃ / r₁₃²

        F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

         r₁₃ = x₃- 0

         r₂₃ = 2 –x₃

We substitute

      k q q / x₃² = k 4q q / (2-x₃)²

      q² (2 - x₃)² = 4 q² x₃²

        4- 4x₃ + x₃² = 4 x₃²

        5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

        x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2  5

        x₃ = [-4 ± 9.80] 10

       X₃₁ = 0.58 m

       X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

7 0
3 years ago
PLEASE HEELP!!!
Ann [662]

Answer:

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8 0
3 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
3 years ago
Tudy the images about geologic time.
Inessa [10]

Answer:

The first flowering plants appeared in the Mesozoic era, not the Paleozoic era.

Explanation:

The Mesozoic era was an era where numerous organisms started to develop in very unique and more advanced ways, both the animals and the plants. In the last period of the Mesozoic, the Cretaceous, the first flowering plants started to appear on the scene. This was revolutionary trait of the plants, and soon these plants started to occupy more and more space and became one of the dominant organisms on the planet. Other important evolution that took place in this period are the appearance of the dinosaurs and the mammals, both becoming the dominant animals on the planet, first the dinosaurs, after that the mammals.

8 0
3 years ago
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