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zimovet [89]
3 years ago
9

Light incident on a polarizer is then passed through a second polarizer. If the polarizer and the analyzer are perpendicular to

each other, what can you say about the light waves emerging from the second polarizer?
a. The light waves are partially blocked.
b. The light waves are completely blocked.
c. The light waves are of the same intensity as the incident light.
d. The light waves have three times the intensity of the incident light.
e. The light waves have half the intensity of the incident light.
Physics
2 answers:
Y_Kistochka [10]3 years ago
7 0
B.

The first polarization will restrict the plane of oscillation for the light, and the second one will block it completely as it is perpendicular to the first.
natima [27]3 years ago
3 0

Answer:

b. The light waves are completely blocked.

Explanation:

When light incident on first polarizer then intensity of light becomes half of its initial intensity after passing through it.

It is because oscillation of light is confined into one plane and light becomes polarized after passing through it

Now another polarizer is placed in front of it at some angle then as per Malus law the intensity of light after passing through that is given by

I = I_0 cos^2\theta

here we know that

\theta = angle between the axis of analyser and polariser axis

here given that analyser and polariser are inclined at 90 degree with each other

so intensity of light received after passing analyser is

I = I_0 cos^290 = 0

so correct answer will be

b. The light waves are completely blocked.

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Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

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R = 2.2+R1,  and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

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3 years ago
A 2 kg object has a specific heat capacity of 1,700 J/(kg \cdot⋅oC)
Nutka1998 [239]

The amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

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  • According to this question, 2 kg object has a specific heat capacity of 1,700J/kg°C and was raised from a temperature of 15 Celsius to 25 Celsius. The heat absorbed is calculated as follows:

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a) The velocity is 2.94m/s

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Explanation:

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Use the equation below to solve for the velocity at 0.30 seconds

vf=vi+at,

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Step 1: Substitute the variables with the knowns

vf=0m/s+(9.8m/s^2)*0.30seconds

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Use the equation below to solve for the displacement at 0.30 seconds

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