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ad-work [718]
3 years ago
15

Ordinary glasses are worn in front of the eye and usually 2.00 cm in front of the eyeball. A certain person can see distant obje

cts well, but his near point is 50.0 cm from his eyes instead of the usual 25.0 cm . Suppose that this person needs ordinary glasses What focal length lenses are needed to correct his vision ?What is their power in diopters?
Physics
1 answer:
Lelu [443]3 years ago
7 0

Answer:

The focal length is 15.549 cm

The power of the lens is 0.0643 D

Solution:

As per the question:

The near point is 50.0 cm

Distance of the glasses from the eyeball, d = 2.00 cm

The near point of a normal human eye is 25 cm

Now,

The image distance, v' = 50.0 - 2.00 = 48.0 cm

The object distance, u' = 25.0 - 2.00 = 23.0 cm

Now, using the Lens maker formula to calculate the focal length:

\frac{1}{f} = \frac{1}{u'} + \frac{1}{v'}

\frac{1}{f} = \frac{1}{48.0} + \frac{1}{23.0} = 0.0643

f = 15.549 cm

Now, the power of the lens in diopters is given by:

P = \frac{1}{f} = \frac{1}{15.549} = 0.0643 D

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Δs / Δt = 2sin2  π  + 3cos 2 π -  ( 2sin π + 3cos π ) / 2 - 1

Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1

Δs / Δt = 6 cm/s

Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s

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The complete question is:

The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1

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2 years ago
How far away is mars?
Elanso [62]

Answer:

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4 years ago
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Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?
shtirl [24]

Answer:

L=6.21m

Explanation:

For the simple pendulum problem we need to remember that:

\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0,

where \theta is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

\omega^{2}=\frac{g}{L},

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There is also an equation that relates the oscillation period and the angular frequeny:

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(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m

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The elements with the largest atomic radii are found in the
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Elements with the largest atomic radius are found in the lower left hand of the periodic table.
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3 years ago
The ability to clearly see objects at a distance but not close up is properly called ________. The ability to clearly see object
melamori03 [73]

Answer:

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     In hyperopia ,people face difficulties  to see close up object , but can see object easily which are at a distance.

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