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34kurt
1 year ago
10

The angle between the axes of two polarizing filters is 45.0^\circ45.0 ​∘ ​​ . By how much does the second filter reduce the int

ensity of the light coming through the first? Select the correct answer 0.750 0.800 0.250 0.500 Your Answer 0.125 Saved 1 of 3 attempts used Part b (1 points) What now is the direction of polarization of the transmitted light? (You can assume the first polarizer defines the angle \theta = 0θ=0.) Select the correct answer
Physics
1 answer:
suter [353]1 year ago
7 0

Answer

given,                                                                      

angle between two polarizing filters = 45°

filter reduce intensity = ?                          

a) I = I₀ Cos² θ                                

here θ = 45⁰                                

I = \dfrac{I_0}{2}                      

intensity of the light is reduced by 0.500

correct answer from the given option D

b) direction of the polarization                    

                        θ = 45°                  

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A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
tekilochka [14]
This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
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The total normal force Fn then is: 
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> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely,  so we can say that the horizontal component is zero: 

Fh = Fph + Fgh 
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> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

 

μs = Fh / Fn

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<span>= 0.47</span>

8 0
2 years ago
Hippos spend much of their lives in water, but amazingly, they don’t swim. manatees, They have, like little very body fat. The d
kenny6666 [7]

Answer:

428.59 N

Explanation:

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Vg\rho_{w}+F_{upward}=mg

F_{upward}=mg- Vg\rho_{w}

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F_{upward}=mg(1-\frac {\rho_{w}}{\rho_{hippo}})

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F_{upward}=1500*9.81*(1-1000/1030)= 428.5922 N

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3 0
2 years ago
An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni
Marysya12 [62]

When air is blown into the open pipe,

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⇒λ=\frac{2L} {n}

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Given L=6m, n=4, solving for λ we get:

λ=\frac{(2)*(6)}{4} =3m

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c=f.λ Or f= \frac{c}{λ}

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8 0
2 years ago
The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

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Explanation:

In order to get a better understanding of this question let us explain some concepts

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We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

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an also this mean that torques acting on the body is balanced

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Therefore we have for each hand = 392.44/2 =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

             Lg = 0.4881 ×  392.44

                  = 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

                  191.22/2 = 95.8 N.

8 0
2 years ago
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