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ad-work [718]
3 years ago
14

Unpolarized light of intensity I=10 falls on two successive polarizer wheels with the angle between the polarizer wheels e-60°.

What is the resultant intensity of the polarized light after leaving the 2nd polarizer? A) (0)/2 B) (10)/4 C) (0)/8 D) (o)/16
Physics
1 answer:
yan [13]3 years ago
5 0

Answer:

I=\frac{10}{4}

Explanation:

A polarizer changes the orientation of the oscillations of a light wave.

I₀ = Intensity of unpolarized light = 10

θ = Angle given to the polarizer = 60°

Intensity of light

I = I₀cos²θ

⇒I = 10cos²60

\\\Rightarrow I=10\times \frac{1}{2}\times \frac{1}{2}\\\Rightarrow I=\frac{10}{4}

So, the after passing through the second polarizer is \mathbf{\frac{10}{4}}

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LekaFEV [45]

The time taken to complete her run is 1.9 hr.

<u>Explanation:</u>

Speed is a scalar quantity and it is defined as the ratio of distance covered to the time taken to cover that distance. As distance is also a scalar quantity, so the directions given in the problem can be ignored. Thus, the distance covered by the jogger is the sum of kilometers given in problem.

Distance covered = 6+5+4 = 15 km

And the speed is given as 8 km/hr.

So the time taken will be ratio of distance to speed.

\text { time }=\frac{\text {distance}}{\text {speed}}=\frac{15}{8}=1.9 \text { hour }

So the jogger will take 1.9 hr to complete her run.

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They carry opposite charge ( one has negative charge and one has positive charge)
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One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor
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Part a)

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Part b)

v = 7.47 m/s

Explanation:

Part a)

(a) when the large-mass object is the one moving initially

So here we can use momentum conservation as the net force on the system of two masses will be zero

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so here we can say

v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}

v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}

v = 16.52 m/s

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}

v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}

v = 7.47 m/s

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If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val
baherus [9]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}

\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \  body \ (v) = 20 \ m/s}

\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \  body \ (u) = 0}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \  body \ ( F)}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

☯ <u>Using 1st equation of motion </u>

\\

\dashrightarrow\:\: \sf{v = u + at}

\\

\dashrightarrow\:\: \sf{20 = 0 + a(4)}

\\

\dashrightarrow\:\: \sf{20 = 4a}

\\

\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}

\\

\dashrightarrow\:\: \sf{a = 5}

\\

☯ <u>Now, Finding the force exerted </u>

\\

\dashrightarrow\:\: \sf{F = ma}

\\

\dashrightarrow\:\: \sf{F = 40 \times 5}

\\

\dashrightarrow\:\: \sf{F = 200 \ N}

\\

☯ <u>Hence</u>, \\

\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}

8 0
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