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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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Answer:

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Explanation:

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A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3
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Explanation:

Initial speed of the rocket, u = 0

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Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

v=u+at

v=25.4\times 3.39=86.10\ m/s    

Let x is the initial position of the rocket. Using third equation of kinematics as :

v^2=u^2+2ax_o

x_o=\dfrac{v^2}{2a}

x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m  

Let x_o is the position at the maximum height. Again using equation of motion as :

v^2-u^2=2a(x-x_o)

Now a=-g and v and u will interchange

u^2=2g(x-x_o)

x=x_o+\dfrac{u^2}{2g}

x=145.92+\dfrac{(86.10)^2}{2\times 9.8}

x = 524.14 meters

Hence, this is the required solution.

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3 years ago
1.Convert 340 cm into m *(answer=0.34m)
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Answer:

<em>1</em><em>.</em><em>for </em><em>the </em><em>first </em><em>one </em><em>100c</em><em>e</em><em>n</em><em>t</em><em>i</em><em>m</em><em>e</em><em>t</em><em>e</em><em>r</em><em>s</em><em> </em><em>make </em><em>1</em><em> </em><em>meter </em><em>therefore</em>

<em>100c</em><em>m</em><em>-</em><em>1</em><em>m</em>

<em>3</em><em>4</em><em>0</em><em>c</em><em>m</em><em>-</em><em>x</em>

<em>3</em><em>4</em><em>0</em><em>/</em><em>100</em>

<em>=</em><em>3</em><em>.</em><em>4</em>

<em>the </em><em>answer </em><em>is </em><em>supposed</em><em> to</em><em> be</em><em> </em><em>3</em><em>.</em><em>4</em><em>,</em><em> maybe</em><em> </em><em>there's</em><em> </em><em>a </em><em>mistake</em><em> </em><em>with </em><em>the </em><em>question</em><em> </em><em>or </em><em>the </em><em>answer</em>

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<em>weight</em><em>=</em><em>7</em><em>5</em><em>×</em><em>9</em><em>.</em><em>8</em>

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<em>I </em><em>hope</em><em> this</em><em> helps</em>

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Answer:

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s;s;

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