<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
Complete Question
A student bikes to school by traveling first dN = 0.900 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.
Similarly, let d⃗ W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form.
Express your answer as two numbers separated by a comma. Be careful with your signs.
Answer:
The value is 
Explanation:
From the question we are told that
The first displacement is 
i.e positive y-axis
The second displacement is 
i.e negative x-axis
The final displacement is 
i.e negative y-axis
Generally dW in component for is
Generally the total displacement of the student is mathematically represented as
Answer:
The answer is 4200 J.
Explanation:
The formula of work done is, W = F×D where F is the force of an object and D is the distance. Then you just substitute the values into the equation :
W = F×D
F = 42N
D = 100m
W = 42 × 100
= 4200 J
Answer
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Explanation:v
Answer:
-------1
Explanation:
beacuse that is what i know
