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qaws [65]
2 years ago
5

How are light waves used to bring far away objects into view and how does the eye translate them?

Physics
1 answer:
adelina 88 [10]2 years ago
5 0

Answer:

When focused light is projected onto the retina, it stimulates the rods and cones. The retina then sends nerve signals are sent through the back of the eye to the optic nerve. The optic nerve carries these signals to the brain, which interprets them as visual images.

Explanation:

Hope it will help u

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Convert 2 days into second​
eduard

Answer:

<em><u>172,000 second </u></em>

<em><u>I'M</u></em><em><u> </u></em><em><u>NOT</u></em><em><u> </u></em><em><u>SURE</u></em><em><u> </u></em><em><u>THAT</u></em><em><u> </u></em><em><u>THIS</u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>RIGHT</u></em><em><u> </u></em><em><u>OR</u></em><em><u> </u></em><em><u>WRONG</u></em><em><u> </u></em><em><u> </u></em><em><u>IF</u></em><em><u> </u></em><em><u>IT'S</u></em><em><u> </u></em><em><u>WRONG</u></em><em><u> </u></em><em><u>THEN</u></em><em><u> </u></em><em><u>SORRY</u></em><em><u> </u></em>

3 0
3 years ago
Read 2 more answers
Please Please Please help on these 3 questions :) 20 POINTS!! - NO LINKS PLEASE
GenaCL600 [577]

Answer:

Catapult on the ground: Normal, gravity

Catapult (I'm assuming launching marshmallow): Reaction of Force Applied

Marshmallow: Force Applied

Explanation:

This is the forces that act on a stationary object and a launched object. The catapult may also experience a force friction if your teacher is taking a more practical sense.

3 0
3 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
3 years ago
A ray in glass arrives at the glass-water interface at an angle of 48° with the normal. The refracted ray, in water, makes a 72°
dimaraw [331]

Answer:

50.4°

Explanation:

Snell's law states:

n₁ sin θ₁ = n₂ sin θ₂

where n is the index of refraction and θ is the angle of incidence (relative to the normal).

When θ₁ = 48°:

n sin 48° = 1.33 sin 72°

n = 1.702

When θ₁ = 37°:

1.702 sin 37° = 1.33 sin θ

θ = 50.4°

3 0
3 years ago
You have created a document that uses the term Frank L. Wright over 30 times. You want to change every occurrence of this term t
fgiga [73]
Find and Replace dialog box is probably the most useful.
3 0
3 years ago
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