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3 years ago

If the person can't accommodate and the glasses is + 2.50d, at which distance will the person see clearly

Physics

2 answers:

bezimeni [28]3 years ago

8 0

Power of the glasses required is given as +2.5 D

this shows that person is having Farsightedness in which he can objects placed far clearly but he is not able to see clearly to the near objects

For normal vision near point of eye is 25 cm

Let say person can see the objects clearly at distance "d" from the eye

now from the formula

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here given that

$\frac{1}{f} = 2.5$

$f = 40 cm$

also we know that

$d_o = 25 cm$

from above equation now

$\frac{1}{d} + \frac{1}{25} = \frac{1}{40}$

$d = 66.6 cm$

So the distance of object must be 66.6 cm

Verdich [7]3 years ago

3 0

The answer would be 40 centimeters

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A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

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Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of the softball, measured in meters.

$y$ - Final height of the softball, measured in meters.

$v_{o}$ - Initial velocity of the softball, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $y = y_{o}$ , $t = 3.56\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the initial velocity of the softball is:

$v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0$

$3\cdot v_{o} -44.132\,m= 0$

$v_{o} = 14.711\,\frac{m}{s}$

The initial velocity of the softball is 14.711 meters per second.

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3 years ago

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Answer:

Explanation:

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Filling in:

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