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3 years ago

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Suppose a flexible, adaptive iol has a focal length of 3.00 cm. how far forward must the iol move to change the focus of the eye

from an object at infinity to an object at a distance of 50.0 cm?

1 answer:

chubhunter [2.5K]3 years ago

8 0

47.00 is your answer.

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Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60

geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar ( $a_{D}$ ), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities ( $v_{D}$ , $\omega_{BD}$ ), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations ( $a_{D}$ , $\alpha_{BD}$ ), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

$v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta$ (1)

$\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta$ (2)

<h3>Accelerations</h3>

$a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma$ (3)

$-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma$ (4)

If we know that $\theta = 60^{\circ}$ , $\gamma = 19.889^{\circ}$ , $r_{BD} = 10\,in$ , $\omega_{AB} = 16\,\frac{rad}{s}$ , $r_{AB} = 3\,in$ and $\alpha_{AB} = 0\,\frac{rad}{s^{2}}$ , then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

$v_{D}+3.402\cdot \omega_{BD} = -41.569$ (1)

$9.404\cdot \omega_{BD} = -24$ (2)

$v_{D} = -32.887\,\frac{in}{s}$ , $\omega_{BD} = -2.552\,\frac{rad}{s}$

<h3>Accelerations</h3>

$a_{D}+3.402\cdot \alpha_{BD} = -445.242$ (3)

$-9.404\cdot \alpha_{BD} = -687.264$ (4)

$a_{D} = -693.867\,\frac{in}{s^{2}}$ , $\alpha_{BD} = 73.082\,\frac{rad}{s^{2}}$

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. $\blacksquare$

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

5 0

2 years ago

You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin

Viktor [21]

Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

W₀ = G m M / Re²

In the calculation of the weight of the person on the 100th floor the only thing that changes is the distance

r = Re + 100 r₀

Where r₀ is the distance between the floors, which is approximately 2.5 m, so the distance is

r = Re + 250

We substitute

W = G m M / r²

W = G m M / (Re + 250)²

The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

W = G m M / Re² (1+ 250 / Re)²

(1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

The value of the expression is

(1 + 250 / Re)⁻² = 1 -2 250 / 6.37 10⁶ -30 (250 / 6.37)² 10⁻¹² + ...

We can see that the quadratic term is very small, which is why we despise it, we substitute in the weight equation

W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

W = Wo (1 - 7.85 10⁻⁵)

We can see that if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

4 0

3 years ago

Read 2 more answers

A dolphin can swim at a constant speed of 12.5 m/s. How

myrzilka [38]

Answer:

$\boxed {\tt 3.6 \ seconds}$

Explanation:

Time can be found by dividing the distance by the speed.

$t=\frac{d}{s}$

The distance is 45 meters and the speed is 12.5 meters per second.

$d= 45 \ m \\s= 12.5 \ m/s$

$t=\frac{45 \ m}{12.5 \ m/s}$

Divide. Note that the meters, or "m" will cancel each other out.

$t=\frac{45 }{12.5 \ s}$

$t=3.6 \ s$

It will take the dolphin 3.6 seconds to swim a distance of 45 meters are 12.5 meters per second.

6 0

3 years ago

Read 2 more answers

The argument against your claim (what the other side would say if they disagreed with your claim.) is:

Vlad1618 [11]

Answer:

a counterclaim

Explanation:

authors purpose is what an author wrote somthing for

opinion is someones thoughts or "side" on a argument

an arguement is a battle of opinions if that makes sense

3 0

3 years ago

Dissolving common salt in water is physical or chemical change?

Ipatiy [6.2K]

Answer: Physical

Explanation: There are no changes being done to the salt. Think of it this way, if you dissolve the salt in water then boil the water completely, you will have the same salt you started with.

8 0

3 years ago