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guapka [62]
3 years ago
7

Unpolarized light with an intensity of 432 W/m^2 passes through three polarizing filters in a row, each of which is rotated 30 d

egrees from the one before it. Find the intensity of the light that emerges at the end
Physics
1 answer:
liraira [26]3 years ago
3 0

Answer:

40.5 W/m²

Explanation:

Intensity of unpolarized light = 432 W/m² = I₀

Intensity of light as it passes through first polarizer

I = I₀×0.5

⇒I = 432×0.5

⇒I = 216 W/m²

Intensity of light as it passes through second polarizer. So, Intensity after it passes through first polarizer is the input for the second polarizer

I = I₀×0.5 cos²30

⇒I = 216×0.75

⇒I = 162 W/m²

Intensity of light as it passes through third polarizer.

I = I₀×0.5 cos²30×cos²60

⇒I = 162×0.25

⇒I = 40.5 W/m²

Intensity of light at the end is 40.5 W/m²

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Write the nuclear reaction equation for the beta decay of Iodine-131
den301095 [7]

Answer:

_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}

Explanation:

In a beta (minus) decay, a neutron in a nucleus turns into a proton, emitting a fast-moving electron (called beta particle) alongside with an antineutrino.

The general equation for a beta decay is:

^A_Z X \rightarrow _{Z+1}^AY+^0_{-1}e+ ^0_0\bar{\nu} (1)

where

X is the original nucleus

Y is the daughter nucleus

e is the electron

\bar{\nu} is the antineutrino

We observe that:

  • The mass number (A), which is the sum of protons and neutrons in the nucleus, remains the same in the decay
  • The atomic number (Z), which is the number of protons in the nucleus, increases by 1 unit

In this problem, the original nucles that we are considering is iodine-131, which is

_{53}^{131}I

where

Z = 53 (atomic number of iodine)

A = 131 (mass number)

Using the rule for the general equation (1), the dauther nucleus must have same mass number (131) and atomic number increased by 1 (54, which corresponds to Xenon, Xe), therefore the equation will be:

_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}

7 0
3 years ago
Read 2 more answers
A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of static friction between t
Wewaii [24]

Answer:

a) a = 6.1 m/s^2

b)  a = 0.98m/s^2

Explanation:

Mass of slab = 40kg

Mass of block = 10kg

Coefficient of static friction (Us)  = 0.60

Kinetic coefficient (UK)  = 0.40

Horizontal force = 100N

The normal reaction from 40kg slab on 10 kg block = 10*9.81

= 98.1N

Static frictional force = Us*R

= 98.1*0.6

= 58.86N

This is less than the force applied

If 10 kg block will slide on the 40 kg slab,  net force = 100 - kinetic force

Kinetic force (Uk*R) = 0.4*98.1

= 39.28N

= 39N

Net force = 100 -39

= 61N

Recall that F = ma

For 10 kg block

a = F/m

a = 61/10

a = 6.1m/s^2

b) Frictional force on 40 kg slab by 10 kg = 98.1*0.4

= 39.24

= 39N

F = ma

a = F/m

For 40kg slab

a = 39/40

a = 0.98m/s^2

3 0
3 years ago
Anders suffered a shock when his electric radio dropped into the bathtub--while anders was taking a bath. Anders argued that he
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The claim Anders is most likely to make is the failure of the manufacturer to warn about such risk.

<h3>What is a Risk?</h3>

This is defined as the possibility of something bad happening and in this case it is electric shock when dropped into the bathtub.

Anders can decide to sue for not warning against risk of electric shock when in contact with water.

Read more about Risk here brainly.com/question/1224221

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2 years ago
When comparing equal volumes of a stronger and weaker acid, what would be observed?
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If you are in plato the answer is B.
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I need help dudes!
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In the late Devonian Era
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