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marusya05 [52]
3 years ago
11

When light is incident normally on the interface between two transparent optical media, the intensity of the reflected light is

given by the expression(equation 1)In this equation S1 represents the average magnitude of the Poynting vector in the incident light (the incident intensity), S’1 is the reflected intensity, and n1 and n2 are the refractive indices of the two media. The fraction of incident intensity is 0.0246 or 2.46 precent for 589-nm light.For light normally incident on an interface between vacuum and a transparent medium of index n is given by S2/S1 = 4n/(n + 1)2 (equation 2). (a) Light travels perpendicularly through a diamond slab, surrounded by air, with parallel surfaces of entry and exit. Apply the tramsmission fraction in equation 2 to find the approximate overall transmission through the slab of diamond, as a percentage. Ignore light relected back and forth within the slab.(b) Light travels perpendicularly through a diamond slab, surrounded by air, with parallel surfaces of entry and exit. The intensity of the transmitted light is what fraction of the incident intensity? Include the effects of light reflected back and forth inside the slab.If someone could answer part (a) and (a) I would gladly appreciate it.
Physics
1 answer:
Alex777 [14]3 years ago
8 0

i need to ask sum so oycixgixigdtdidit

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Its tangential speed is constant although its velocity is changing. As the object changes direction, it results in a changing of positive and negative signs of the velocity. Although, the magnitude of the velocity (speed) is not changing.

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An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency o
kodGreya [7K]

Answer:

Part A

The efficiency of the engine is 10%

Part B

The change in internal energy is 300 J

Part C

The change in volume is 1 m³ which is one cubic meter

Explanation:

Part A

Efficiency is defined as the ratio of energy output to energy input;

The given parameters are;

The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

The amount of energy produced by the engine = 1,000 J

Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

∴ The energy output to the car by the engine = F × d = 100 N × 10 m = 1,000 J

The energy input from the engine = The energy produced by the engine = 10,000 J

The efficiency of the engine = (The energy output)/(The energy input)= 1,000J/10,00J = 0.1

The efficiency in percentage = 0.1 × 100 = 10 %

The efficiency of the engine = 10%

Part B

The amount of heat added to the substance, ΔQ = 1,000J

The amount of work the substance does on the atmosphere, W = 700 J

The change in internal energy, ΔU is given as follows;

ΔQ = W + ΔU

∴ ΔU = ΔQ - W

For the substance, we have;

The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

Part C

The work done by the piston, W = 1,000 J

The pressure, in the piston, P = 1,000 Pa = constant

The work done by the piston in a constant pressure process, W = P × ΔV

Where;

W = The work done

P = The constant pressure applied

ΔV = The change in volume = V₂ - V₁

V₂ = The final volume

V₁ = The initial volume

∴The change in volume ΔV = W/P = 1,000 J/(1,000Pa) = 1 m³

The change in volume ΔV = 1 m³

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Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

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There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

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Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

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