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egoroff_w [7]
3 years ago
10

Experiments show that the ground spider uses one of its several pairs of eyes as a polarization detector. In fact, the two eyes

in this pair have polarization directions that are at right angles to one another. Suppose linearly polarized light with an intensity of 845 W/m2 shines from the sky onto the spider, and that the intensity transmitted by one of the polarizing eyes is 263 W/m2. For this eye, what is the angle between the polarization direction of the eye and the polarization direction of the incident light
Physics
1 answer:
irina [24]3 years ago
4 0

Answer:

The angle between the polarization direction of the eye and the polarization direction of the incident light = 56.1°

Explanation:

For polarization, the relationship between the intensity of transmitted rays (I), intensity of incident rays (I₀) and the angle between the polarization direction of the eye and the polarization direction of the incident light (θ) is given as

I = I₀ cos²θ

I = intensity of transmitted rays = 263 W/m²

I₀ = intensity of incident rays = 845 W/m²

θ = ?

263 = 845 cos²θ

cos²θ = (263/845) = 0.3112

Cos θ = √0.3112 = 0.5579

θ = cos⁻¹ (0.5579)

θ = 56.1°

Hope this Helps!!!

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a block is pushed up a frictionless 40 incline. if the initial velocity after the push is 5.00 m/s. how far along the incline do
Lana71 [14]

Answer:

d=1.982m, t=1.019s

Explanation:

There are different approaches we can take to solve this problem. You could either solve this by using conservation of energy or by taking a kinematic approach. I'll solve this by using kinematics. So, the very first thing we need to do in order to solve this is do a drawing of the situation so we can analyze it better. (See attached picture).

So, since we are talking about an inclined plane, we can see that the force of gravity is being split into an x and y components if we incline the axis of coordinates. Taking this into account we can see that:

\sum F_{x}=ma_{x}

Since there is no friction in our system, then the only force acting upon the box is the force of gravity, or weight. Since we are taking the upwards direction as the positive direction of movement, we can say that the force of gravity is excerting a negative influence on our box, so this acceleration will be negative, so our sum of forces will now look like this:

-mg sin(40^{o})=ma

we can cancel the masses out so we can see that:

a=-g sin(40°)

a=-9.81m/s^{2} sin(40^{o})

We have now enough information to solve our problem.

we can take the following equation to find the distance the block travels up the incline:

x=\frac{V_{f}^{2}-V_{0}^{2}}{2a}

we know the final velocity must be zero, so we can use the provided data to solve our formula:

x=\frac{(0)^{2}-(5m/s)^{2}}{2(-9.81m/s^{2})sin 40^{o}}

which yields:

x=1.982m

In order to find the time it takes for the block to return to its original position we can use the following formula:

x=V_{0}t+\frac{1}{2}at^{2}

since x=0 is the starting point we can use that to solve our equation:

0=5t+\frac{1}{2}(-9.81sin 40^{o})t^{2}

which simplifies to:

0=5t-4.905t^{2}

which can now be solved for t

t(5-4.905t)=0

t=0                and          5-4.905t=0

t=0                 and         t=\frac{5}{4.905}=1.019

so the time it takes the block to return to its original position is

t= 1.019s

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3 years ago
What are the units for acceleration?
Anni [7]
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8 0
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Read 2 more answers
A 500g air-track glider attached to a spring with spring constant 10 N/m is sitting at rest on a frictionless air track. A 250 g
d1i1m1o1n [39]

To solve this problem it is necessary to apply the conservation equations of the moment for an inelastic impact or collision. In turn, it is necessary to apply the equations related to the conservation of potential energy and kinetic energy.

Mathematically this definition can be expressed as

m_1v_1+m_2v_2=(m_1+m_2)v_f

Where,

v_{1,2} = Initial velocity of each object

m_{1,2} =Mass of each object

v_f =Final velocity

Our values are given as

m_1 = 0.25kg\\m_2 = 0.5kg\\v_1 = 1.2m/s\\v_2 = 0m/s

Replacing we can find the value of the final velocity, that is

0.750v_f = 0.25*1.20+0.5*0

v_f = 0.4m/s

From the definition of the equations of simple harmonic motion the potential energy of compression and equilibrium must be subject to

KE_{compressed}+PE_{compressed} =KE_{equilibrium} +PE_{equilibrium}

Since there is no kinetic energy due to the zero speed in compression, nor potential energy at the time of equilibrium at the end, we will have to

0+\frac{1}{2}KA^2 = \frac{1}{2}(m_1+m_2)v_f^2+0

Re-arrange to find A

A = \sqrt{\frac{m_1+m_2}{k}v_f}

A = \sqrt{\frac{0.25+0.5}{10}(0.4)}

A = 0.11m

Finally, the period can be calculated through the relationship between the spring constant and the total mass, that is,

T = 2\pi \sqrt{\frac{m_1+m_2}{k}}

T = 2\pi \sqrt{\frac{0.5+0.25}{10}}

T = 1.7s

6 0
3 years ago
A ski gondola is connected to the top of a hill by a steel cable of length 600 m and diameter 1.2 cm . As the gondola comes to t
Gemiola [76]

Answer:

v = 66.7 m/s

Explanation:

Given that,

The length of steel cable, L = 600 m

Diameter = 1.2 cm

It is observed that it took 18 s for the pulse to return.

The time taken to cover 600 m will be :

t = T/2

t = 9 s

Let v be the of the pulse. We know that,

v=\dfrac{L}{t}\\\\v=\dfrac{600}{9}\\\\v=66.7\ m/s

So, the speed of the pulse is equal to 66.7 m/s.

6 0
3 years ago
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