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Galina-37 [17]
3 years ago
6

A nearsighted person wants to see an apple that is 7 meters away but can only clearly see objects that are at most 62 cm away fr

om her eyes. Eyeglasses made of diverging corrective lenses can help her to see the apple clearly. If her glasses are 2.5 cm in front of her eyes, what should be the power of these corrective lenses to allow her to see the apple?
Physics
1 answer:
natulia [17]3 years ago
6 0

Answer:

The power of the corrective lenses is 3.162 D.

Explanation:

Given that,

Object distance = 70 cm

Image distance = 62 cm

Distance between eyes and glasses = 2.5 cm

Eyeglasses made of diverging corrective lenses can help her to see the apple clearly

So now ,

Object distance from glass =70-2.5 = 67.5 cm

Image distance from glass = 62-2.5 = 59.5 cm

We need to calculate the focal length

Using formula for focal length

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

\dfrac{1}{f}=-\dfrac{1}{59.5}-\dfrac{1}{67.5}

\dfrac{1}{f}=-\dfrac{508}{16065}\ cm

We need to calculate the power of lens

Using formula of power

P=\dfrac{100}{f}

P=-\dfrac{508}{16065}\times100

P=-3.162 \ D

Negative sign shows the lens is diverging.

Hence, The power of the corrective lenses is 3.162 D.

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In this experiment, the ______ was intentionally manipulated. This was the independent variable.
Marizza181 [45]

Answer:

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3 years ago
Final exam answers for physical science apex?
Tju [1.3M]
First table:

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5ftx\frac{0.305m}{1ft}  = 1.525mx\frac{100cm}{1m} =152.5cm

then we convert the remaining 2in into cm and add it up:
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Another way to do this is to convert the ft to in and add it up to the remaining 2 in then convert to cm. The answer would be more or less the same. 

2. For weight, we will convert lbs to kg. 

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3. For distance we will convert mi to km. 

1.1mix\frac{1.61km}{1mi} = 1.8km

4. The unit mph is miles per hour and we will want to convert that into kph or kilometers per hour. You will use the same method that you did above to get it. We do not need to convert hours anymore, so it will stay as is.

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4 0
3 years ago
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
A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo a
Mekhanik [1.2K]

Answer:

1.17 m

Explanation:

From the question,

s₁ = vt₁/2................ Equation 1

Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.

Given: v = 343 m/s, t = 0.0115 s

Substitute into equation 1

s₁ = (343×0.0115)/2

s₁ = 1.97 m.

Similarly,

s₂ = vt₂/2.................. Equation 2

Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo

Given: v = 343 m/s, t₂ = 0.0183 s

Substitute into equation 2

s₂ = (343×0.0183)/2

s₂ = 3.14 m

The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁

s₂-s₁ =  (3.14-1.97) m = 1.17 m

7 0
3 years ago
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