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4 years ago

An incident ray parallel to the axis and one through the focal point reflect and converge at the image.

Physics

2 answers:

Grace [21]4 years ago

3 0

Answer:

The answer is true.

Explanation:

A ray coming from the object hits the surface of a mirror and then passes through focus point after reflection. Since the image is formed by the actual meeting of the reflected rays therefore it is called as real image. the direction of arrow will be upward.

Same is true if the object is placed at center of curvature the coming ray along the parallel axis passes through the focus and bends downward. The point where the two reflecting rays will meet is the point of image. And the direction of image arrow will be downward. In this case image is formed between F and C.

melomori [17]4 years ago

3 0

Answer:

TRUE

Explanation:

In mirrors, all rays of light incidented on a plane surface (mirror) and parallel to the principal axis converges at the focus of the mirror after reflection.

When forming the image of an object placed in front of a mirror (for example curved mirrors), there are two steps:

1. , all rays of light incidented on a plane surface (mirror) and parallel to the principal axis converges at the focus of the mirror after reflection.

2. Light rays from the object passes through the centre of curvature undeflected and reflect back on the same line of incidence

The two reflected rays from formed will intersect at a point and this point is where the image will be formed.

Based on the conclusion above the statement in question is TRUE i.e

an incident ray parallel to the axis and one through the focal point reflect and converge at the image.

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Ber [7]

Answer:

Explanation:

I think you got it right

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3 years ago

Read 2 more answers

A block with mass of 10 kg is on a frictionless surface. One hand on the left side of the block is pushing it to the right. A se

igomit [66]

Answer:

$W_2=-12J$

Explanation:

The work of force 2 will be given by the vectorial equation $W_2=F_2.d$ . We know the value of $F_1$ and have information about its movement, which relates to the net force $F=F_1+F_2$ .

About this movement we can obtain the acceleration using the equation $v_f^2=v_i^2+2ad$ . Since it departs from rest we have $a=\frac{v_f^2}{2d}$ .

And then using Newton's 2dn Law we can obtain the net force F=ma, thus we will have $F_2=F-F_1=ma-F1=\frac{mv_f^2}{2d}-F_1$

And we had the work done by force 2 as:

$W_2=F_2.d=\frac{mv_f^2}{2}-F_1d$

(The sign will be given algebraically since we take positive the direction to the right.)

With our values:

$W_2=\frac{(10kg)(2m/s)^2}{2}-(8N)(4m)=-12J$

Another (shorter but maybe less intuitive way for someone who is learning) way of doing this would have been to say that the work done by both forces would be equal to the variation of kinetic energy:

$W_1+W_2=K_f-K_i=K_f=\frac{mv_f^2}{2}$

Which leads us to the previous equation straightforwardly.

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3 years ago

Which of the following situations has more kinetic energy than potential energy?

san4es73 [151]

Answer: I think its C

Explanation:

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3 years ago

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Electrons on the outermost shell/energy level

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3 years ago

A truck covers 40.0 m in 7.10 s while smoothly slowing down to a final velocity of 2.75 m/s.. (a) Find the truck's original spee

ser-zykov [4K]

D = vo t - at²/2
v = vo - at
40 m = vo · 7.10 - a · ( 7.10 )²/2
40 m = 7.10 vo - a· 50.41/2
40 m = 7.10 vo - 25.205 a
2.75 m/s = vo - 7.10 a ⇒ a = ( vo - 2.75 )/ 7.10
40 = 7.10 vo - 25.205 · ( vo - 2.75 ) / 7.10
40 = 7.1 vo - 3.55 vo + 9.7625
3.55 vo = 30.2375
vo = 30.2375 : 3.55 = 8.5176 m/s
Answer: the truck`s original speed is 8.5176 m/s.

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4 years ago