Answer:
The near point of an eye with power of +2 dopters, u' = - 50 cm
Given:
Power of a contact lens, P = +2.0 diopters
Solution:
To calculate the near point, we need to find the focal length of the lens which is given by:
Power, P = 
where
f = focal length
Thus
f = 
f = 
= + 0.5 m
The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.
Now, by using lens maker formula:
where
u = object distance = 25 cm = 0.25 m = near point of a normal eye
u' = image distance
Now,
Solving the above eqn, we get:
u' = - 0.5 m = - 50 cm
Answer:
x = 2.044 m
Explanation:
given data
initial vertical component of velocity = Vy = 2sin18
initial horizontal component of velocity = Vx = 2cos18
distance from the ground yo = 5m
ground distance y = 0
from equation of motion
solving for t
t = 1.075 sec
for horizontal motion
x = 2cos18*1.075
x = 2.044 m
To solve this problem it is necessary to apply the concepts related to the capacitance in the disks, the difference of the potential and the load in the disc.
The capacitance can be expressed in terms of the Area, the permeability constant and the diameter:
Where,
= Permeability constant
A = Cross-sectional Area
d = Diameter
Potential difference between the two disks,
V = Ed
Where,
E = Electric field
d = diameter
Q = Charge on the disk equal to 
Through the value found and the expression given for capacitance and potential, we can define the electric charge as
Re-arranging the equation to find the diameter of the disks, the equation will be:
Replacing,
Therefore the diameter of the disks is 0.03m
Answer:
τ = 132.773 lb/in² = 132.773 psi
Explanation:
b = 12 in
F = 60 lb
D = 3.90 in (outer diameter) ⇒ R = D/2 = 3.90 in/2 = 1.95 in
d = 3.65 in (inner diameter) ⇒ r = d/2 = 3.65 in/2 = 1.825 in
We can see the pic shown in order to understand the question.
Then we get
Mt = b*F*Sin 30°
⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in
Now we find ωt as follows
ωt = π*(R⁴ - r⁴)/(2R)
⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)
⇒ ωt = 2.7114 in³
then the principal stresses in the pipe at point A is
τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)
⇒ τ = 132.773 lb/in² = 132.773 psi
The image will form in the vicinity of F. Its nature will be small and inverted
