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sergejj [24]
3 years ago
11

Two polarizers are oriented at 68 ∘ to one another. unpolarized light falls on them. part a what fraction of the light intensity

is transmitted?
Physics
2 answers:
kykrilka [37]3 years ago
7 0

Answer:

Fraction=0.0701

or

Fraction=7.01%

Explanation:

Apply Malus Law

I₁=I₀/2

I₂=I₁cos²θ

I₂=(I₀/2)cos²(68)

I₂=I₀(0.1403/2)

Fraction of light is given as

I₂/I₀=0.1403/2=0.0701

To get in %

Fraction=0.0701×100

Fraction=7.01%

Radda [10]3 years ago
4 0
Your answer is.07 hope this helped 
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Multiply each side by 'volume' :    (density) x (volume) = (mass)

Divide each side by 'density' :                         Volume = (mass) / (density)
 

5 0
3 years ago
Read 2 more answers
Hi! Can somebody please help?
dezoksy [38]

Answer:

Diagram A will reach the top first.

Explanation:

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7 0
3 years ago
A stone is thrown vertically upwards with a speed of 30.0 m/s.
matrenka [14]

a)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = final position of stone = 20.0 meters

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = ?

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 30² + 2 (- 9.8) (20 - 0)

v = 22.5 m/s


b)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = maximum height gained

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = 0 m/s

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

0² = 30² + 2 (- 9.8) (Y - 0)

Y = 46 m



6 0
3 years ago
A series L-R-C circuit consists of a 226 Ω resistor, a 27.4 mH inductor, a 11.55 µF capacitor, and an AC source of amplitude 15
DanielleElmas [232]

Answer: 363 Ω.

Explanation:

In a series AC circuit excited by a sinusoidal voltage source, the magnitude of the impedance is found to be as follows:

Z = √((R^2 )+〖(XL-XC)〗^2) (1)

In order to find the values for the inductive and capacitive reactances, as they depend on the frequency, we need first to find the voltage source frequency.

We are told that it has been set to 5.6 times the resonance frequency.

At resonance, the inductive and capacitive reactances are equal each other in magnitude, so from this relationship, we can find out the resonance frequency fo as follows:

fo  = 1/2π√LC = 286 Hz

So, we find f to be as follows:

f = 1,600 Hz

Replacing in the value of XL and Xc in (1), we can find the magnitude of the impedance Z at this frequency, as follows:

Z = 363 Ω  

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