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2 years ago

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Unpolarized light of intensity Io passes through six successive Polaroid sheets each of whose axis makes a 60 ∘ angle with the p

revious one. What is the intensity of the transmitted beam?

2 answers:

dolphi86 [110]2 years ago

5 0

Answer:

$0.00048828125I_0$

Explanation:

$I_0$ = Unpolarized light intensity

$\theta$ = Angle = 60°

Intensity of light through first polarizer

$I_1=\dfrac{I_0}{2}$

Intensity of light through nth polarizer

$I=I_1(cos^2\theta)^{n-1}$

when

n = 6

$I_6=\dfrac{I_0}{2}(cos^2(60))^{6-1}\\\Rightarrow I_6=0.00048828125I_0$

The intensity of light after passing through the sixth polarizer is $0.00048828125I_0$

arlik [135]2 years ago

3 0

Answer:

0.0004883 Io

Explanation:

Let Io be the intensity of unpolarised light.

Intensity of light coming out from first polaroid.

I1 = Io / 2

According to law of Malus, the intensity of light coming from the n th polaroid is given by

$I_{n}= I_{1}\left (Cos^{2}\theta \right )^{n-1}$

where, n be the number of polaroid and θ be the angle between two polaroid.

$I_{6}= \frac{I_{0}}{2}\left (Cos^{2}60 \right )^{6-1}$

So, I6 = 0.0004883 Io

Thus, the intensity of light coming out from the sixth polaroid is 0.0004883 Io.

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Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

PE = (2.5)(98)(14) and

PE = 340 J

b. $KE=\frac{1}{2}mv^2$ so

$KE=\frac{1}{2}(2.5)(14)^2$ and

KE = 250 J

c. TE = KE + PE so

TE = 340 + 250 and

TE = 590 J

d. PE at 8.7 m:

PE = (2.5)(9.8)(8.7) and

PE = 210 J

e. The KE at the same height:

TE = KE + PE and

590 = KE + 210 so

KE = 380 J

f. The velocity at that height:

$380=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(380)}{2.5} }$ so

v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

590 = KE + PE and

PE = (2.5)(9.8)(11.6) so

PE = 280 then

590 = KE + 280 so

KE = 310 then

$310=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(310)}{2.5} }$ so

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h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

$v=v_0+at$ and

26 = 0 + 9.8t and

26 = 9.8t so the time at 26 m/s is

t = 2.7 seconds. Now we use that in the equation for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in the time the object was at 26 m/s:

Δx = 0t + $\frac{1}{2}(-9.8)2.7)^2$ so

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i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

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you have a 200 kg bag being lifted with a block and tackle. If you pull with 100 newtons what is the MA of the system?

Korvikt [17]

Answer:

The mechanical advantage of the system is equal to 19.62

Explanation:

The ratio of the force produced by a machine to the force applied to it is called mechanical advantage. In other words it is the ratio of output force to the input force.

In this problem mass=200kg

applied force=100N

input force=100N

output force= $mg=200 \times 9.8=196N.$

mechanical advantage $MA= \frac{1960}{100}=19.6$

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Answer:

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Rate of mass accumulation inside vessel = mass flow in - mass flow out

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taking p as density

$d \frac{dQ}{dt} = pq_i_n$

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q(in) is the volume flow rate coming in

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But,

dQ = Adh

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A = area of liquid surface at time t

h = height from bottom at time t

A = πr²

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Hence,

$\pi( \frac{h}{2} )^2\frac{dh}{dt} =q_i_n$

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