If a nearsighted person has a far point df that is 3.50m from the eye, what is the focal length f1 of the contact lenses that th
e person would need to see an object at infinity clearly?
If a farsighted person has a near point that is 0.600m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350m from the person's eyes?
If a farsighted person has a near point that is 0.600m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350m from the person's eyes?
1 answer:
Answer:
f1= -350cm or -3.5m
f2= 22.1cm or 0.221m
Explanation:
A person is nearsighted when the person's far point is less than infinity. A diverging lens is normally used to correct this eye defect. A diverging lens has a negative focal length as seen in the solution attached.
Farsightedness is when a person's near point is farther than 25cm. This eye defect is corrected using a converging lens. The focal length of a converging lens is positive. This is evident in the solution attached. The near point is also referred to as the least distance of distinct vision.
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Answer:
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Explanation:
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Answer:
(a)Therefore the highest altitude attained by the object is =576 ft .
(b)Therefore the object takes 6 sec to fall to the ground.
Explanation:
Initial velocity: Initial velocity is a velocity from which an object starts to move.
u is usually used for notation of initial notation.
Final velocity: Final velocity is a velocity of an object after certain second from starting.
The final velocity is denoted by v.
Acceleration: The difference of final velocity and initial velocity per unit time
The S.I unit of acceleration is m/s².
(a)
Given that u= 128 ft\sec and g = 32 ft/sec².
At highest point the velocity of the object is 0 i.e v=0
Since the displacement is opposite to the gravity.
Therefore acceleration( a)= -g = -32 ft/sec².
To find the time this to happen we use the following formula
Here v=0
⇒0=128+(-32) t
⇒32t=128
⇒t = 4 sec
To determine the height we use the following formula
⇒s= 256 ft
Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .
(b)
At the highest point the velocity of the object is 0.
so u=0. a=g= 32 ft/sec² [ since the direction of gravity and the displacement are same] s= 576 ft
To determine the time to fall we use the following formula
⇒t=6 sec
Therefore the object takes 6 sec to fall to the ground.