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il63 [147K]
3 years ago
9

Write a sentence to describe how an image is formed on retinawhen looking at distant objects.​

Physics
1 answer:
beks73 [17]3 years ago
7 0

Answer:

An image is formed on the retina with light rays converging most at the cornea and upon entering and exiting the lens. Rays from the top and bottom of the object are traced and produce an inverted real image on the retina. The distance to the object is drawn smaller than scale

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What does it mean to say the mass is conserved during a physical change ?
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Mater doesn't just appear or disappeared. Chemical elements are still there just the connections and how it combines changes.

So what goes into your chemical eqation must still exist after the change.
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4 years ago
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Write the products for the following neutralization reactions
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Answer:

answer 3

Explanation:

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If I bought 59 kids off the black market in a secret door in the stalls and I sell 21 kids then ask for 81 more then 12 off the
larisa [96]

You will be left with 106 kids

<h3>Meaning of word problem</h3>

A word problem can be defined as a mathematical problem that is written in word or written in a sentence format.

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2 years ago
Under which condition does Ohm's law apply? a. The current must be constant b. The power must be constant c. The temperature mus
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3 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
Margaret [11]

Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

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3 0
4 years ago
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