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vredina [299]
2 years ago
11

What is the size of the image on the retina of a size object 1.5 cm, placed at a distance (120 cm) away? Take the .lens-to-retin

a distance to be 2 cm
Physics
1 answer:
Mariulka [41]2 years ago
6 0

Answer: 0.025 cm

Explanation:

The image distance must match the distance between the lens and the retina for clear vision. Consequently, the image distance is v = 2 cm

The magnification of the lens is given by

m = v/u = h'/h

  • Where h' is the height of the image

v/u = h'/ h

h' = ( v/u ) × h

h' = ( 2cm /-120cm ) × 1.5cm

h' = - 0.025cm

Therefore, the height of the image is 0.025 cm

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Natalka [10]

Answer:

475 m , 950 m

Explanation:

Let l be the length of the side perpendicular to the barn.

1900-2l = length of the side parallel to the barn

Area A= l( 1900-2l)

A= 1900l-2l^2

now, the maximum value of l ( the equation being quadratic)

l_max= -b/2a

a= 2

b=1900

l_max= -1900/4= 475 m

then 1900-2l= 1900-2×(475) = 950 m

So, the dimensions that maximize area are

950 and 475

Now. A_max = -2( l_max)^2+1900×l_max

A_max=  -2(475)^2+1900×475

A_max= 451250 m^2

or, 475×950 = 451250 m^2

6 0
4 years ago
An isobar connects areas of the same _____ _____. An isotherm connects areas of the
WINSTONCH [101]

Hi there!

So, I believe an <em>isobar</em> connects areas of the same atmospheric pressure, and an <em>isotherm</em> connects areas of the same temperature. Hope this helped :)

8 0
3 years ago
just want to double-check that you understand a practical consequence of the expansion of the universe. Light reaches us from a
Daniel [21]

Answer:

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<em></em>

Explanation:

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speed of light in m/years is = (3 x 10^8)/(60 x 60 x 24 x 365) = 9.4608 x 10^15 m/year

distance = speed x time

therefore, the distance of this light = 10 x 10^9 x 9.461 x 10^15 = <em>9.4608 x 10^25 m</em>

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Answer:

Explanation:

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