<h2>Question: </h2>
The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens?
Answer:
29.96cm
Explanation:
Using the corrective lens, the image should be formed at the front of the eye and be upright and virtual.
Now using the lens equation as follows;
-------------(i)
Where;
f = focal length of the lens
v = image distance as seen by the lens
u = object distance from the lens
From the question;
v = -151cm [-ve since the image formed is virtual]
u = 25cm
Rewrite equation (i) to have;
Substitute the values of v and u into the equation;
f = 29.96cm
The focal length should be 29.96cm
The distance covered is simply the length of the entire trip, which is 12m + 16m, or 28m.
The displacement is the distance from the starting point to the ending point along with the direction of the net motion. The dog walks 12m east then 16m west, so its resultant displacement is 4m west.
Answer:
18 m/s
Explanation:
15 m/s + 3 m/s = 18 m/s
Answer:
1. The story happens at Masayahin Senior High School,during his first class in grade 11.
2. The transitional divices are used is At first,then,after a year, in the end.
3. I entered the class and jasper offered me a seat.
4. A story.
5. in chronological orde
Answer:3.95 m/s
Explanation:
Given
mass of object 
radius of circle 
initial Position 
angular displacement 
8.95 radian can be written as
i.e. Particle is at first quadrant with 
(c)velocity is 
