To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.
From the law of Malus intensity can be defined as

Where
Angle From vertical of the axis of the polarizing filter
Intensity of the unpolarized light
The expression for the intensity of the light after passing through the first filter is given by

Replacing we have that


Re-arrange the equation,

Re-arrange to find \theta





The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°
Explanation:
Equilibrium position in y direction:
W = Fb (Weight of the block is equal to buoyant force)
m*g = V*p*g
V under water = A*h
hence,
m = A*h*p
Using Newton 2nd Law

Hence, T time period
T = 2*pi*sqrt ( h / g )
Answer:
H_w = 2.129 m
Explanation:
given,
Width of the weir, B = 1.2 m
Depth of the upstream weir, y = 2.5 m
Discharge, Q = 0.5 m³/s
Weir coefficient, C_w = 1.84 m
Now, calculating the water head over the weir




now, level of weir on the channel
H_w = y - H
H_w = 2.5 - 0.371
H_w = 2.129 m
Height at which weir should place is equal to 2.129 m.
Answer:
3.0 cm
Explanation:
We can solve this problem by using the mirror equation:

where
f is the focal length of the mirror
p is the distance of the object from the mirror
q is the distance of the image from the mirror
In this problem we have:
f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)
p = 3.0 cm is the distance of the object from the mirror
Therefore, the distance of the image is:

And the positive sign means that the image is real.
(The second part of the exercise is just the description of the image of the first exercise).
How much work in J does the string do on the boy if the boy stands still?
<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>
<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>
<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>
<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>