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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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A motorcycle has a constant acceleration of 3.49 m/s2. Both the velocity and acceleration of the motorcycle point in the same di
Vilka [71]

Answer:

(a)2.865 s

(b)2.865 s

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We are given that

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Final speed,v=39 m/s

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t=\frac{39-29}{3.49}=2.865 s

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A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in
miv72 [106K]

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is \mu_s =  0.60

The value for static friction is \mu_k =  0.70

Explanation:

From the question we are told that

The mass of the clock is m  =  95 \  kg

The first horizontal force is F _s  =  560 \  N

    The second horizontal force is    F _k  =  650  \  N

Generally the static frictional force is equal to the first  horizontal force

So

     F _s  =  m  *  g  *  \mu_s

=>   560  =  95  *  9.8  *  \mu_s

=>    \mu_s =  0.60

Generally the kinetic frictional force is equal to the second horizontal force

So

      F _k  =  m  *  g  *  \mu_k

      650 =  95  *  9.8  *  \mu_k

     \mu_k =  0.70

3 0
3 years ago
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