These three members of the Nile gas family have one property in common because they are gases at room temperature. That is they

Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second

vampirchik [111]

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

$I' = \frac{I_0}{2} cos^2 \theta$

Where

$\theta =$ Angle From vertical of the axis of the polarizing filter

$I_0 =$ Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

$I = \frac{I_0}{2}$

Replacing we have that

$I = \frac{370}{2}$

$I = 185W/m^2$

Re-arrange the equation,

$I'= \frac{I_0}{2}cos^2\theta$

Re-arrange to find \theta

$cos^2\theta = \frac{2I'}{I_0}$

$cos^2\theta = \frac{2*138}{370}$

$\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})$

$\theta = 0.5282rad$

$\theta = 30.27\°$

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

4 0

3 years ago

A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top a

Marysya12 [62]

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

$-m*\frac{d^2y}{dt^2} = Fb - W\\\\-m*\frac{d^2y}{dt^2} = p*g*(h+y)*A - A*h*p*g\\\\-A*h*p*\frac{d^2y}{dt^2} = y *p*A*g\\\\\frac{d^2y}{dt^2} + \frac{g}{h} * y =0$

Hence, T time period

T = 2*pi*sqrt ( h / g )

4 0

3 years ago

A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl

Alex

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

$Q = C_w BH^{3/2}$

$H = (\dfrac{Q}{C_wB})^{2/3}$

$H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}$

$H = 0.371\ m$

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.

7 0

3 years ago

Use the diagram below to answer the following question:

d1i1m1o1n [39]

Answer:

3.0 cm

Explanation:

We can solve this problem by using the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

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:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.5}-\frac{1}{3.0}=\frac{1}{3.0}\\\rightarrow q=3.0 cm$

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).

5 0

2 years ago

A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .

sp2606 [1]

How much work in J does the string do on the boy if the boy stands still?

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 
--------------------------------------... 
How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? 

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. 

If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: 
4.5N + (boys negative acceleration * mass) = total force1 
work = total force1 x 11 meters 
--------------------------------------... 

How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? 

answer: same as above only reversed: 
4.5N - (boys negative acceleration * mass) = total force2 
work = total force2 x 11 meters

6 0

3 years ago