Ask question

Ask question

All categories

Hunter-Best [27]

2 years ago

14

Discuss the path that light takes through the human eye

2 answers:

amm18122 years ago

8 0

The cornea is shaped like a dome and bends light to help the eye focus. Some of this light enters the eye through an opening called the pupil (PYOO-pul). The iris (the colored part of the eye) controls how much light the pupil lets in. Next, light passes through the lens (a clear inner part of the eye).

WITCHER [35]2 years ago

4 0

From the cornea, the light passes through the pupil. The iris, or the colored part of your eye, controls the amount of light passing through. From there, it then hits the lens. This is the clear structure inside the eye that focuses light rays onto the retina.

You might be interested in

Based on the model, which statement best explains why point A is

Vladimir79 [104]

It’s D.Earths northern hemisphere is tilted toward the sun

6 0

3 years ago

Read 2 more answers

A 0.0208 m diameter coin rolls up a 18.0◦ inclined plane. The coin starts with an initial angular speed of 56.0 rad/s and rolls

anastassius [24]

Answer:

h = 0.0259 m

Explanation:

given,

diameter of the cone = 0.0208 m

radius,r = 0.0104 m

angle of inclination,θ = 18°

initial angular velocity, ω_i = 56 rad/s

final angular velocity ,ω_f = 0 rad/s

height, h = ?

Rotational kinetic energy

$KE_r = \dfrac{1}{2}I\omega^2$

Moment of inertia of coin

$I = \dfrac{1}{2}MR^2$

so,

$KE_r = \dfrac{1}{4}MR^2\omega^2$

Transnational Kinetic energy

$KE_t = \dfrac{1}{2}Mv^2$

v = r ω

$KE_t = \dfrac{1}{2}MR^2\omega^2$

now,

using conservation energy

Kinetic energy of the coin is converted into the potential energy

KE_r + KE_t = PE

$\dfrac{1}{4}MR^2\omega^2 + \dfrac{1}{2}MR^2\omega^2 = Mgh$

$\dfrac{3}{4}R^2\omega^2=gh$

$\dfrac{3}{4}\times 0.0104^2\times 56^2=9.8\times h$

h = 0.0259 m

Vertical height gain by the coin is equal to 0.0259 m

7 0

4 years ago

Can someone help label these?

seropon [69]

A. reactants
B. subscript
C. coefficient
D. products

7 0

2 years ago

For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this

ikadub [295]

Answer:

no idea

Explanation:

7 0

4 years ago

How much kinetic energy is in a 0.145kg baseball thrown at 35.8 m/s ?

Mazyrski [523]

K= (1/2) m v^2
so 0.5*0.145*<span>1281.64=92.92J of KE</span>

4 0

4 years ago