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4 years ago

What happens to light waves at the interface between different media?

Physics

1 answer:

nevsk [136]4 years ago

6 0

Answer:

Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.

Explanation:

Hope this helps.

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Tyler has learned that potential energy is energy stored. Tyler's teacher asks the students to come up with a demonstration of p

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D.A child at the top of a slide

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4 years ago

Read 2 more answers

magine two carts, one with twice the mass of the other, that are going to have a head-on collision. In order for the two carts t

scoray [572]

Answer:

Twice as fast

Explanation:

Solution:-

- The mass of less massive cart = m

- The mass of Massive cart = 2m

- The velocity of less massive cart = u

- The velocity of massive cart = v

- We will consider the system of two carts to be isolated and there is no external applied force on the system. This conditions validates the conservation of linear momentum to be applied on the isolated system.

- Each cart with its respective velocity are directed at each other. And meet up with head on collision and comes to rest immediately after the collision.

- The conservation of linear momentum states that the momentum of the system before ( P_i ) and after the collision ( P_f ) remains the same.

$P_i = P_f$

- Since the carts comes to a stop after collision then the linear momentum after the collision ( P_f = 0 ). Therefore, we have:

$P_i = P_f = 0$

- The linear momentum of a particle ( cart ) is the product of its mass and velocity as follows:

m*u - 2*m*v = 0

Where,

( u ) and ( v ) are opposing velocity vectors in 1-dimension.

- Evaluate the velcoity ( u ) of the less massive cart in terms of the speed ( v ) of more massive cart as follows:

m*u = 2*m*v

u = 2*v

Answer: The velocity of less massive cart must be twice the speed of more massive cart for the system conditions to hold true i.e ( they both come to a stop after collision ).

8 0

3 years ago

Pressure in the abdomen

Vilka [71]

Answer:

Uh, what? Is there more?

Explanation:

3 0

3 years ago

Samuel applies a horizontal force of 35.0 N to a sleigh over a distance of 1.50 m along a level surface. Calculate the work done

Ronch [10]

Answer:

52.5 J

Explanation:

Work done (W) is the product of the force (F) applied on a body and the distance (s) moved in the direction of the force.

i.e W = F × s

It is a scalar quantity and measured in Joules (J).

Given that: F = 35.0 N and s = 1.50 m, then;

W = F × s

W = 35.0 × 1.5

= 52.5 J

Therefore, the work done on the sleigh by Samuel is 52.5 J.

5 0

3 years ago

A water strider bug is supported on the surface of a pond by surface tension acting along the interface between the water and th

Alik [6]

Answer:

minimum interface length = 1.36 mm

Explanation:

given data

weight of the bug = $10^{-4}$ N

solution

we will apply here Surface Tension formula that is

Surface Tension ,σ = Force ÷ length ........................1

and we consider here surface tension for water is 7.34 × $10^{-2}$ N/m

so that here minimum interface length needed to support the bug is

minimum interface length = Force ÷ σ

minimum interface length = $10^{-4}$ ÷ 7.34 × $10^{-2}$

minimum interface length = 1.36 mm

4 0

3 years ago