Answer:
d.The wavelength of light and the size of the aperture
Explanation:
<em>The resolution power of an optical system is the smallest distance between two points that the device can distinguish clearly.</em>
It has the following relationship:

where:
r = minimum resolvable distance
n = numerical aperture
= wavelength of the light used for viewing
From above mathematical equation it is clear that:
- Smaller the wavelength better the resolving power
- Larger the aperture better the resolution
(Note, that smaller the value of "r" the more finer details of the image visible through the device.)
Answer:
Explanation:
Remark
In general, these 3rd class levers are very inefficient. Because the force distance is smaller than the load distance, you need to pull upward with more force that the weight of the load. So whatever the load is, the force is going to be much greater.
The distances are always measured to the pivot unless you are asked something specific otherwise.
Givens
F = ?
weight = 6N
Force Distance = F*d = 0.5 m
Weight Distance =W*d1 = 2 m
Formula
F*Fd = W*Wd
Solution
F*0.5 = 6 * 2 Divide by 0.5
F = 12/0.5
F = 24 N upwards
Given:
density of air at inlet, 
density of air at inlet, 
Solution:
Now,

(1)
where
A = Area of cross section
= velocity of air at inlet
= velocity of air at outlet
Now, using eqn (1), we get:

= 1.14
% increase in velocity =
=114%
which is 14% more
Therefore % increase in velocity is 14%
Answer:
a =( -0.32 i ^ - 2,697 j ^) m/s²
Explanation:
This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.
Break down the speeds in two moments
initial
v₀ₓ = v₀ cos θ
v₀ₓ = 5.25 cos 35.5
v₀ₓ = 4.27 m / s
= v₀ sin θ
= 5.25 sin35.5
= 3.05 m / s
Final
vₓ = 6.03 cos (-56.7)
vₓ = 3.31 m / s
= v₀ sin θ
= 6.03 sin (-56.7)
= -5.04 m / s
Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order
a = (
- v₀) /t
aₓ = (3.31 -4.27)/3
aₓ = -0.32 m/s²
= (-5.04-3.05)/3
= -2.697 m/s²