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svetoff [14.1K]
3 years ago
10

A converging lens focuses a set of light rays entering the lens on one side parallel to its axis to a single focal point on the

opposite side of the lens. if the principal rays emitted from a single point on an object all intersect at a single point after passing through a converging lens, then this intersection point helps in determining the location and size of a real image of the object. the most useful principal rays for converging lenses are the following:
Physics
1 answer:
Vladimir79 [104]3 years ago
7 0

1)The P ray is directed parallel to the axis of the lens. It emerges from the lens directed toward the focal point on the side of the lens opposite the object.


2) The M ray is directed toward the midpoint of the lens. It emerges from the lens unchanged in direction.


3) The F ray is directed toward the focal point on the same side of the lens as the object. It emerges from the lens directed parallel to the lens axis.


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An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
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To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

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The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

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Here,

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\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

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The  volume flow rates for ∆P is 6.81m³/s .

<h3>What is pressure?</h3>

The amount of force applied on perpendicular to the surface of an object per unit area. The unit of it is pascal.

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As there are two tubes then the height of tube 1 is equal to height of tube two .

P1-P2=1/2p(Vd²-Vl²)

The flow rate of liquid is  A1V1=A2V2 .

rest is attached in image

to learn more about Pressure click here brainly.com/question/12971272

#SPJ4

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