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tatiyna
3 years ago
8

A woman wears bifocal glasses with the lenses 2.0 cm in front of her eyes. The upper half of each lens has power-0.500 diopter a

nd corrects her far vision so that she can focus clearly on distant when looking through that half. The lower half of each lens has power +2.00 diopters and corrects her near vision when she looks through that half What are the far point and near point of her eyes?
Physics
1 answer:
saveliy_v [14]3 years ago
8 0

Answer:

q = -2 m  and  q = -0.5 m

Explanation:

For this exercise we must use the equation of the optical constructor

        1 / f = 1 / p + 1 / q

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

Let's start with the far vision point, in this case the power of the lens is

        P = -0.5D

power is defined as the inverse of the focal length in meter

      f = 1 / D

      f = -1 / 0.5

      f = - 2m

the object for the far vision point is at infinity p = infinity

     1 / f = 1 / p + i / q

      1 / q = 1 / f - 1 / p

      1 / q = -1/2 - 1 / ∞

       q = -2 m

The sign indicates that the image is on the same side as the object

Now let's lock the near view point

D = +2.00 D

f = 1 / D

f = 0.5m

the near mink point is p = 25 cm = 0.25 m

        1 / f = 1 / p + 1 / q

        1 / q = 1 / f - 1 / p

        1 / q = 1 / 0.5 - 1 / 0.25

        1 / q = -2

        q = -0.5 m

the sign indicates that the image is on the same side as the object in front of the lens

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A helicopter flies 25 km north, 5 km east, then 5 km S, then 15 km W. What is the resultant displacement and direction of the he
ch4aika [34]

Answer:

Explanation:

Plotting the original location of the helicopter before it flies 25 km north, it would be at the origin, (0,0) then after it flies north, the y vertex gains 25 points, so it would be (0,25)

After it flies east, the x coordinate gains 5 points, so it would now be (5,25)

After it flies south, the y coordinate loses or is subtracted by 5 points. so it would now be (5,20)

After flying west, the x coordinate loses 15 points. So the final vertex would be at (-10,20)

East = Right

West = Left

South= Down

North = Up

I used mainly mathematical methods by adding and subtracting the x and y coordinate values, but this could be graphed easily since I gave the coordinates just incase!

Hope this helps!

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2 years ago
8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h
pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

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So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

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(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0
3 years ago
a runner starts from rest and has an acceleration of 3 m/s^2. How fast is she running after 1.1 seconds
grigory [225]
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Since you're given acceleration and time, just plug the values into the equation. 

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Solve that equation, and remember your velocity should be in m/s.
8 0
2 years ago
When an electron in a certain excited energy level in a one-dimensional box of length 2.00 Å makes a transition to the ground st
Fiesta28 [93]

Answer:

Calculate the wavelength associated with an electron with energy 2000 eV.

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3 years ago
One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given dista
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Answer:

(a) 0.94 m/s²

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Explanation:

(a)

From Newton's equation of motion,

S = ut + 1/2gt²......................... equation 1

Making g the subject of equation 1

g =( S - ut)/t² ........................ equation 2

Where  s = distance ( m), u = initial velocity (m/s), t = time (s), g = acceleration due to gravity (m/s²)

From the question, S = 12.02 m, t = 3.58 s, u= 0 ( at rest),

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g = {12.02 -(0×3.58)}/3.58²

g = (12.02)/12.82

g = 0.94 m/s²

∴ The acceleration due to gravity on the planet = 0.94 m/s²

(b) g (planet)/g (earth) = 0.94/9.80

     g (planet) = 0.096 g (earth).

The acceleration due to gravity of the planet in terms of the earth g  is

g (planet) = 0.096g

5 0
3 years ago
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