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bixtya [17]
2 years ago
12

A beam of unpolarized light with a power of 1.0 mW passes through two ideal linear polarizers in series. An ideal polarizerabsor

bs100% of the light perpendicular to itspolarization axis, and none of the light parallel to itspolarization axis. What should the angle between the polarization axes of the two polarizers be if you want the power emerging from the second polarize?
Physics
1 answer:
Mashutka [201]2 years ago
3 0

Answer:

the axes of the two polarizers are aligned, the angle between this axis must be zero

Explanation:

Polarizers are anisotropic systems where absorption is selective in each axis, in this case they indicate that on an axis they let 100% pass and perpendicular to this axis no light passes.

  The first polarizer defines a direction of light that of its axis, it is said that the light is polarized, when this light reaches the second polarized, so that it can pass the axis must be in the direction of the polarized beam.

In summary, the axes of the two polarizers are aligned, the angle between this axis must be zero

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A girl is bouncing on a trampoline where is her gravitational potential energy a maximum and where is her kinetic energy maximum
Stella [2.4K]

Answer:

When you jump down, your kinetic is converted to potential energy of the stretched trampoline. The trampoline's potential energy is converted into kinetic energy, which is transferred to you, making you bounce up. At the top of your jump, all your kinetic energy has been converted into potential energy. Right before you hit the trampoline, all of your potential energy has  been converted back into kinetic energy. As you jump up and down your kinetic energy increases and decrease.

7 0
3 years ago
A cannonball is fired vertically upwards at 100.0 m/s a) How long will it take to return to the cannon? b) what is it's maximum
V125BC [204]
Answer:
a) 20s
b) 500m

Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.

To find time, we apply the UARM formula:

v final = (a x t) + v initial

Replacing the values gives us:

0 = (-10 x t) + 100

-100 = -10t

t = 10s

It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur

So 10s going up and another 10s going down:

10x2 = 20s

b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:

Δy = (1/2)(a)(t^2) + (v initial)(t)

Replacing the values gives us:

Δy = (1/2)(-10)(10^2) + (100)(10)

= (-5)(100) + 1000

= -500 + 1000

= 500 m

Hope this helps, brainliest would be appreciated :)
7 0
2 years ago
On a frictionless surface how much force is necessary to accelerate a 0.49 kg object to the left at 4.8 m/s2?
Alenkasestr [34]

Answer:

<h2>2.35 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

force = 0.49 × 4.8 = 2.352

We have the final answer as

<h3>2.35 N</h3>

Hope this helps you

8 0
2 years ago
Which equation can be used to model simple harmonic motion
mixer [17]

Answer:x(t)= Acos(wt)

Explanation:

According to Newton's 2nd law,a particle of mass m acted on by a force is given by:Fs=-kx

Where x is displacement from equilibrium

K = spring constant

Therefore X(t) = Acos(2pit/T)

X(t)= Acos(wt)

8 0
3 years ago
A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

Normal force (N) = mg CosΘ

μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

7 0
3 years ago
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