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dezoksy [38]
4 years ago
7

How did the lens, cornea, and retina work together to make human vision possible?

Physics
1 answer:
Anvisha [2.4K]4 years ago
5 0
First light is passed through the lens and then is focused by the cornea and lens onto a thin layer or tissue called the retina and when the light hits the retina tiny cells collect the light signals and converts them converts them into electrical signals which are then sent into the optic nerve and into the brain and are then processed into the images that we see
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A 2.00 kg object is attached to a spring and placed on frictionless, horizontal surface. Ahorizontal force of 18.0 N is required
velikii [3]

Answer:

6 rad/s

Explanation:

In a spring the angular frequency is calculated as follows:

\omega=\sqrt{\frac{k}{m} }

where \omega is the angular frequency, m is the mass of the object in this case m=2kg, and k is the constant of the spring.

To calculate the angular frequency, first we need to find the constant k which is calculated as follows:

k=\frac{F}{x}

Where F is the force: F=18N, and x is the distance from the equilibrium position: x=0.25m.

Thus the spring constant:

k=\frac{18N}{0.25m}

k=72N/m

And now we do have everything necessary to calculate the angular frequency:

\omega=\sqrt{\frac{k}{m} }=\sqrt{\frac{72N/m}{2kg} }=\sqrt{36}

\omega=6rad/s

the angular frequency of the oscillation is 6 rad/s

7 0
4 years ago
A horizontal line on a velocity/time graph shows ____ acceleration.
Novay_Z [31]

Answer:

Zero (0)

Explanation:

A horizontal line will be no movement of negative of positive acceleration

6 0
3 years ago
Read 2 more answers
Which of the following is considered a calm area close to the equator resulting in little to no wind?
solong [7]
<h2>Right answer: Doldrums</h2>

These are also called zones of equatorial calm and it is due a climatic phenomenon that is placed near the Earth equator, attributed to the soft winds, that are called calm winds as well; accompanied by systems of abundant rains and heat.

In this area periods of great calm occur when the winds virtually disappear completely, trapping the sailing ships for long periods (days or weeks). This is why the term <em>doldrum</em> became popular as a colloquial expression in the eighteenth century, to refer to "<em>the caprice of the wind that slows down the navigation to sail". </em>


The zone is located in the place where two trade winds meet, this means the trade winds of the northern hemisphere <u>converge</u> with those of the southern hemisphere, that is why this region is related to the <u>intertropical convergence zone</u>.


5 0
4 years ago
A small particle has charge -2.00 uC and mass 1.50×10-4 kg. It moves from point A, where the electric potential is V(A) = 200 V,
Svet_ta [14]

Answer:

v_b=6.13 m/s..

Explanation:

Since no external force is acting on the system.

Therefore, Total energy remains constant before and after.

So, Total energy of system= energy due to potential applied+kinetic energy

T.E=qV_a+\dfrac{1}{2}mv_a^2=qV_b+\dfrac{1}{2}mv_b^2\\\dfrac{1}{2}mv_b^2=\dfrac{1}{2}mv_a^2+q(V_a-V_b)\\

(Here v=velocity ,V=potential ,q=charge and m=mass).

Putting values .

We get,  v_b=6.13 m/s..

At point B charged particle is moving faster as compared to point A.

Hence, it is the required solution.

5 0
4 years ago
A solid sphere of diameter D = 26 cm has a charge of Q = 4 nano-coulombs uniformly distributed on it. Calculate the magnitude of
kozerog [31]

Answer:

E = 4.83  N/ C

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere:

E= \frac{k*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

Where:

K: coulomb constant (N*m²/C²)

a: sphere radius (m)

Q:  Total sphere charge (C)

r : Distance from the center of the sphere to the region where the electric field is calculated (m)

Equivalences

1nC=10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=4nC=4 *10⁻⁹C

D = 26 cm = 26*10⁻²m = 0.26m

a = D/2 = 0.13m

r= R+a = 2.6 m+ 0.13m = 2.73m

Problem development

Magnitude of the electric field at r = 2.73m from the center of the sphere  

r>a , We apply the Formula (1) :

E= \frac{k*Q}{r^{2} }

E= \frac{9*10^{9}*4*10^{-9} }{2.73^{2} }

E= 4.83 N/ C

3 0
3 years ago
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