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xxMikexx [17]
3 years ago
5

An object is placed in front of a concave mirror, between the center of curvature of the mirror and its focal point, as shown in

the diagram below. Three light rays are traced, along with their corresponding reflected rays. Which statement below best describes the image formed?
A. The image is real.

B. The image is right-side-up.

C. The image is reduced.

D. No image is formed.

Physics
1 answer:
weeeeeb [17]3 years ago
6 0
<h2>Answer: The image is real</h2>

A concave mirror has a reflective surface that is curved inwards.  This type of mirrors reflect the light making it converge in a focal point therefore they are used to focus the light. This occurs because the light is reflected with different angles, since the normal to the surface varies from one point to another of the mirror.

Now, the formation of the image will depend on the following conditions:

-<u>If the object is at a distance greater than the focal distance</u>, a <u>real</u> and <u>inverted image</u> is formed that may be larger or smaller than the object.  

<u>-If the object is at a distance smaller than the focal distance</u>, a <u>virtual  upright image</u> is formed that is larger than the object.  

As we can see in the figure attached, <u>the object is at a distance greater than the focal distance</u>, this means<u> the image is real</u>. This can be seen in the point where the rays converge.

In addition, this is an special case, <u>the object is between the center of curvature of the mirror and its focal point</u>, which means the image produced is also <u>inverted and larger than the object.</u>

<u></u>

Therefore, the correct option is A.

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The velocity of Satellite A is 2% greater than velocity of satellite B.

The given parameters;

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The forces acting on the Satellites are given as follows;

F_c = \frac{mv^2}{r} \\\\F_g = \frac{GMm}{r^2} \\\\\frac{v^2}{r}  = \frac{GM}{r^2}\\\\v^2 = \frac{GM}{r} \\\\v^2 r = GM\\\\v_1^2 r _1 = v_2^2 r_2\\\\v_A^2r_A = v_B^2 r_B\\\\(\frac{v_A}{v_B} )^2 = \frac{r_B}{r_A} \\\\\frac{v_A}{v_B}  = \sqrt{\frac{r_B}{r_A} } \\\\\frac{v_A}{v_B}  = \sqrt{\frac{(800,000\  + \ 6.4 \times 10^6}{(500,000\  + \ 6.4 \times 10^6} )} \\\\\frac{v_A}{v_B}  =  1.02 \\\\v_A = 1.02 \ v_B

v_A = v_B( 100\% \ + 2\%)\\\\v_A = 100\%v_B \ + \ \ 2\% v_B\\\\v_A = v_B \ \ + \ 2\% v_B

Thus, the velocity of Satellite A is 2% greater than velocity of satellite B.

Learn more about velocity of satellite here: brainly.com/question/13981089

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Answer:

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<u>So rearrange the formula to find the force</u>

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. A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 30.0 m
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