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viva [34]
3 years ago
15

As light waves enter the eye, which three structures do they pass through first?

Physics
1 answer:
lesya692 [45]3 years ago
4 0
As light enters to the eye, it would pass first through three (3) structures or main layers/tunics:
1. Fibrous tunic - outermost layer of the eye. Consists of cornea and sclera which gives the eye white color and protect the inner parts of the eye.
2. Vascular tunic - middle layer of the eye. Consists of iris and choroid which gives the dark color of the eye and inhibits disorderly reflections inside the eye.
3. Nervous tunic - inner layer of the eye. Consist of retina which is responsible for vision
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Stand on a bathroom scale on a level floor, and the reading on the scale shows the gravitational force on you, mg. If the floor
kari74 [83]

Answer:

<em>The angle introduces an error on the mesure of the weight</em>

Explanation:

<u><em>Weight and Normal Forces</em></u>

When an object is resting on a horizontal surface, its weight is directed downwards and the normal force has the same magnitude and opposite direction, i.e. directed upwards. When some angle α exists between the surface and the horizontal plane, the scale keeps 'feeling' the Normal force, but it's not equal to the weight anymore, but to the perpendicular component of the weight to the surface where the scale is placed. It can be found that the component of the weight is m.g.cosα

If, for example \alpha=10^o, the real measure is

m.g.cos 10^o=0.985(mg)

That is, 98.5% of the correct weight. So it's important to measure weight on horizontal surfaces

7 0
3 years ago
The resolution of a telescope is ultimately limited by the diameter of its objective lens or mirror. A typical amateur astronome
slava [35]

Answer:

 θ =  3.19  arc second

Explanation:

The resolution of a telescope is given by the rayleigh criterion, which establishes that two objects are separated if the principal maximum of diffraction of one of them coincides with the first minimum of diffraction of the second object, based on this the solution is given by the first diffraction minimum, the a slit is

        a sin θ = m λ

with m = 1

in the case of circular apertures the equation must be found in polar coordinates, therefore a numerical constant is introduced

        a sin θ = 1.22 λ

Angles are measured in radians and in these experiments they are small

        sin θ = θ

       θ= 1.22  λ  / a

in this case a = 6.09 in, the wavelength is wrong = 550 10⁻⁹ m which is the maximum resolution of the human eye

l

et's reduce the magnitudes to the SI system

        d = 6.09‘  2.54 10⁻-2 m / 1 inch = 15.4686 10-2 m

let's calculate

       θ = 1.22 550 10-9 / 15.468 10-2

       θ = 15.5 10⁻⁶ rad

       rad = 2.06 105 s

       θ = 15.5 10⁻⁶ rad  2.06 105s/ 1 rad

       θ =  3.19   s

     

3 0
3 years ago
A particular baseball pitcher throws a baseball at a speed of 39.1 m/s (about 87.5 mi/hr) toward home plate. We use g = 9.8 m/s2
Reika [66]
There is no acceleration in the horizontal direction (just g in the vertical), so we can use v = d/t, where v is velocity, d is distance and t is time. We can solve for time like so: t = d/v, we can plug in numbers (v is 39.1m/s completely in the horizontal direction, so no need to break it down with sin's and cos's, just plug it in) and we get t = (16.6m)/(39.1 m/s) = 0.42 s. Keep in mind it wouldn't fall far enough vertically to hit home plate (though we don't know the ball's initial height anyway), but would be in the air just above it. Cheers!
6 0
2 years ago
A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follo
cestrela7 [59]

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration (a_c) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration (a_c) is given by:

a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s

The velocity (v) is given by:

v = ωr;  where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

4 0
2 years ago
A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin
anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

\Delta L = L_0 \alpha \Delta T

Where,

L_0 = Initial Length

\alpha = Thermal expansion coefficient

\Delta T = Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C

103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C

The coefficient of thermal expansion given is

\alpha = 1.1*10^{-5}/\°C

The initial length would be,

L_0 = 565m

Replacing we have to,

\Delta L = L_0 \alpha \Delta T

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = 0.355m

This means that the building will be 35.5cm taller

3 0
3 years ago
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