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RoseWind [281]
3 years ago
13

The adjustments that your eyes make as they look from objects near to objects far away or from objects far away to objects close

r can serve as cues to depth in and of themselves. This source of information about depth is known as______.
Physics
1 answer:
Hatshy [7]3 years ago
6 0

Answer:

Oculomotor depth cues is based on two things convergence and accommodation,for the eyes to see near by things within 25cm near point we look inwards at this point the muscles holding the lens tighten up to see near by things unlike when we see far things we tend to look outwards with our eye muscles relaxing.

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Might it be possible to explain the interaction of the rod and pieces of paper as a gravitational interaction? please explain wh
mina [271]

It is not possible  to explain the interaction of the rod and pieces of paper as a gravitational interaction.

<h3>What is Gravitational interaction?</h3>

This is defined as the interaction between a particle or body resulting from their mass. This type of interaction is usually weak and occurs in all distances possible.

It is not gravitational interaction, because the rod attracts paper only against the gravitational force of the earth and there is no attraction between both bodies under a different condition.

This is therefore the reason why it is not possible to explain the interaction of the rod and pieces of paper as a gravitational interaction.

Read more about Gravitational interaction here brainly.com/question/25624188

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5 0
2 years ago
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bija089 [108]
C. Ptolemy argued that earth was the center of the universe
7 0
3 years ago
A fisherman casts his lure at an angle of 33 degrees above the horizontal. The lure reaches a maximum height of 2.3 m. Assuming
omeli [17]

Answer:

12.32 m/s

Explanation:

Using the formula of maximum height of a projectile,

H = u²sin²Ф/2g................... Equation 1

Where H = maximum height, u = initial velocity, Ф = angle of projection, g = acceleration due to gravity

make u the subject of the equation

u = √(2Hg/sin²Ф)............ Equation 2

Given: H = 2.3 m, Ф = 33°, g = 9.8 m/s²

Substitute into equation 2

u = √[(2×2.3×9.8)/sin²33°]

u =√ [45.08/(0.545)²]

u = 45.08/0.297

u = √(151.785)

u = 12.32 m/s

3 0
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Answer:

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What is center of mass?
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