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wel
3 years ago
12

A 80 mW laser beam is polarized horizontally. It then passes through two polarizers. The axis of the first polarizer is oriented

at 30∘ from the horizontal, and that of the second is oriented at 60∘ from the horizontal.
Physics
1 answer:
inn [45]3 years ago
4 0

The E(incident) for the first  polarizer is E(incident)=-76.1930384332

The E(incident) for the second  polarizer is E(incident)=12.340115991

<u>Explanation:</u>

<u>Solving the problem:</u>

<u>Given</u>

Transmitted data = 80 mW

First polarizer= 30 degree

Second polarizer= 60 degree

We have the formula,

E (transmitted) = E (incident)cos()

<u>1. Through the first polarizer</u>

E (transmitted) = E (incident)cos()

80=E(incident) cos(60 degree)     (according to law subtract the given degree with 90 degree).

Then,

E(incident)=E(transmitted) cos()

E(incident) =80×cos 60 degree

E(incident)=-76.1930384332

The E(incident) for the first  polarizer is E(incident)=-76.1930384332

<u>2. Through the first polarizer</u>

We have the formula,

E (transmitted) = E (incident)cos()

80=E(incident) cos(30 degree)     (according to law subtract the given degree with 90 degree).

Then,

E(incident)=E(transmitted) cos()

E(incident) =80×cos 30 degree

E(incident) =12.340115991

The E(incident) for the second  polarizer is E(incident)=12.340115991

<u></u>

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Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

From the question,

We are to calculate the power each student would have to climb the flight of stairs.

Power can be calculated using the formula

P = \frac{F \times d}{t}

Where

P is Power

F is the force

d is the distance

and t is the time

NOTE: The weight of the students represent the force

  • For student 1

F = 700 N

d = 4 m

t = 6 s

∴ P = \frac{700 \times 4}{6}

P = 467 W

  • For student 2

F = 650 N

d = 4 m

t = 6 s

∴ P = \frac{650 \times 4}{6}

P = 433 W

Hence, Student 1 would have a power 467 W and student 2 would have a power of 433 W. The correct option is the fourth option - Student 1 would have 467 W, and Student 2 would have 433 W of power.

Learn more here: brainly.com/question/18801566

3 0
2 years ago
A 1000-kilogram car traveling with a velocity of 20. meters per second decelerates uniformly at -5.0 meters per second2 until it
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Answer:

-20000 kgm/s

Explanation:

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Mathematically, impulse can be expressed as

I = m(v-u).............. Equation 1.

Where I = impulse applied to the car to bring it to rest, m = mass of the car, u = initial velocity of the car, v = final velocity of the car.

Given: m = 1000 kg, u = 20 m/s, v = 0 m/s ( to rest)

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3 years ago
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Answer:

a) 400.4Joules

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Work is said to be done if the force applied to an object cause the object to move through a distance

Workdone = Force × Distance

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Force = 76N

Distance= 5.2m

Work done = 77 × 5.2

Work done = 400.4Joules

b) If the force is exerted at an angle of 41°

Work done = Fdsin theta

Work done = 77(5.2)sin41

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6 0
3 years ago
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Temka [501]

Answer:

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Explanation:

I dunno if this is along the lines of an answer you're looking for, but hope this helps :)

7 0
3 years ago
Can someone please help me with this physics question? I'm desperate!
Lelu [443]

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

3 0
2 years ago
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