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Gala2k [10]
2 years ago
12

I need help ASAP. This is for 15 points

Physics
1 answer:
kramer2 years ago
7 0

Answer:

Latitude :

runs: east to west

measures : distances north and south of the equator

Longitude :

runs : north to south

measures : the distance east or west of the Prime Meridian

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It must be a virtual image, because this is the only kind of image it can produce.

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What is the angle of reflection?
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Answer: the answer is d

Explanation:

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3 years ago
Sarah, who lives in Australia notices that the length of the shadow of the flagpole is getting shorter every day when measured a
Allisa [31]

Answer:

Time zone is one important factor in difference in location and this in turn affects the result of the resolution and rotation of shadow produced from the sun or other illumination.

Therefore someone at a place might see a clear large shadow due to shinny sun reflection and another a small or dull Shadow at same time if the intensity of the sun or lighting source is going down.

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3 0
3 years ago
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

        \frac{1}{f}  = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = \frac{h'}{h} = - \frac{q}{p}

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

           \frac{1}{q} = \frac{1}{f }  - \frac{1}{p}

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0
3 years ago
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