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Ber [7]
3 years ago
7

Why does straightening the curve not change the equation that represents the graph

Physics
2 answers:
3241004551 [841]3 years ago
5 0
Your question may go along with a picture that you're not showing us.

Just reading the question, I can't think of any curved graph that you could
straighten without changing the equation that represents it.
Sphinxa [80]3 years ago
3 0
<span>You cant change the curve of a graph with out changing the equation</span>
You might be interested in
The normal eye, myopic eye and old age
yanalaym [24]

Answer:

1)    f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

        \frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}

        f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

           1 / f = 1/15 + 1 / 1.5

           1 / f = 0.733

            f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

         f’₀ / f = 1.5 / 1.36

         f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

          a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

          tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

          sin θ = y / f

where f is the distance of the lens (eye)

           y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

           y / f = 1.22 λ / D

           y = 1.22 λ f / D

where D is the diameter of the eye

          D = 2R₀

          D = 2 0.1

          D = 0.2 cm

           

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

         

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

      \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

       \frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}

       \frac{1}{f}= 0.7066

        f = 1.415 cm

therefore the diffraction is

        y = 1.22  550 10⁻⁹  1.415  / 0.2

        y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

       d = 2y

       d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

       \frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}

        \frac{1}{f}= 0.733

         f = 1.36 cm

diffraction is

        y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

        y = 4.56 10-6 m

the diffraction diameter is

        d_myope = 2y

         d_myope = 9.16 10-6 m

         \frac{d_{normal}}{d_{myope}} = 9.49 /9.16

        \frac{d_{normal}}{d_{myope}} =  1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0
3 years ago
Select the correct answer.
VLD [36.1K]

Answer:O – H

Explanation:

8 0
2 years ago
A Boeing 747 ""Jumbo Jet"" has a length of 59.7 m. The runway on which the plane lands intersects another runway. The width of t
Reika [66]

Answer:

1.7 seconds

Explanation:

To clear the intersection, the total distance to be covered = 59.7 + 25 =84.7m

first we need to find the initial speed to just enter the intersection by using the third equation of motion

v^2 - u^2 = 2*a*s

45^2 - u^2 = 2 * -5.7 * 84.7

u^2 = 45^2 +965.58

u^2 = 2990.58

u = 54.7 m/s

Now for time we apply the first equation of motion

v-u =a * t

t = (v-u)/a = (45 - 54.7)/-5.7 = 1.7seconds

8 0
3 years ago
A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of
Dafna1 [17]

Answer:

<em>-</em><em> </em><em>p</em><em>e</em><em>n</em><em>d</em><em>u</em><em>l</em><em>u</em><em>m</em><em> </em><em>u</em><em>s</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>d</em><em>e</em><em>t</em><em>e</em><em>r</em><em>m</em><em>i</em><em>n</em><em>e</em><em> </em><em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em>

Explanation:

<u>b</u><u>y</u><u> </u><u>c</u><u>a</u><u>l</u><u>c</u><u>u</u><u>l</u><u>a</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>n</u><u>e</u><u>e</u><u>d</u><u> </u><u>t</u><u>o</u><u> </u><u>u</u><u>n</u><u>d</u><u>e</u><u>r</u><u>s</u><u>t</u><u>a</u><u>n</u><u>d</u><u> </u><u>f</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>p</u><u>r</u><u>o</u><u>b</u><u>l</u><u>e</u><u>m</u>

<em>t</em><em>h</em><em>e</em><em> </em><em>l</em><em>e</em><em>n</em><em>g</em><em>t</em><em>h</em><em> </em><em>i</em><em>s</em><em> </em><em>m</em><em>e</em><em>a</em><em>s</em><em>u</em><em>r</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>b</em><em>e</em><em> </em><em>2</em><em>s</em>

<em>t</em><em>h</em><em>e</em><em> </em><em>l</em><em>e</em><em>n</em><em>g</em><em>h</em><em>t</em><em> </em><em>i</em><em>s</em><em> </em><em>a</em><em>l</em><em>s</em><em>o</em><em> </em><em>2</em>

<em>2</em><em> </em><em>o</em><em>b</em><em>t</em><em>a</em><em>i</em><em>n</em><em>e</em><em>r</em><em> </em><em>o</em><em>n</em><em> </em><em>i</em><em>n</em><em>v</em><em>i</em><em>s</em><em>t</em><em>i</em><em>g</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>m</em><em>o</em><em>s</em><em>t</em><em> </em><em>n</em><em>e</em><em>a</em><em>r</em><em>l</em><em>y</em><em> </em><em>i</em><em>s</em><em> </em><em>8</em><em>.</em>

5 0
2 years ago
The presence of which element would indicate that a star is going through a high-mass star life cycle as opposed to a low-mass s
Delicious77 [7]
Celestial bodies in the universe like the stars, gain their energy by nuclear fusion. This is a nuclear reaction that emits radiation by joining subatomic particles together to yield another new element. This cause by instability of certain elements due to their high neutron-to-proton ratio. The most stable element there is, is Fe-26. Elements lighter than Fe-26 are most likely to undergo nuclear fusion (combining), while elements heavier than Fe-26 are most likely to undergo nuclear fission (breaking). 

So that is how the Sun gains its energy. It is very abundant in hydrogen, such that hydrogen undergoes nuclear fusion. Two protons from two hydrogen atoms combine at very very high temperatures to form a Helium atom. Therefore, a high-mass star life is very abundant in Hydrogen, while a low-mass star life is very abundant in Helium.
7 0
3 years ago
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