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3 years ago

Explain how a single piece of polarizing film can be used to adjust the intensity of a beam of light..

Physics

1 answer:

masya89 [10]3 years ago

6 0

Answer:

Sunlight and almost every other form of natural and artificial illumination produces light waves whose electric field vectors vibrate in all planes that are perpendicular with respect to the direction of propagation.

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Birdman is flying horizontally at a

zavuch27 [327]

Answer:

X=92.49 m

Explanation:

Given that

u= 21 m/s

h= 97 m

Time taken to cover vertical distance h

h= 1/2 g t²

By putting the values

97 = 1/2 x 10 x t² ( g = 10 m/s²)

t= 4.4 s

The horizontal distance

X= u .t

X= 21 x 4.4

X=92.49 m

3 0

3 years ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$

this means, for the angular momentum measure from the rod:

$\frac{d\vec{L}_{rod}}{dt} = 0$

that means :

$\vec{L}_{rod} = constant$

So, the magnitude of initial angular momentum is :

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)$

but the angle is 90°, so:

$| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|$

$| \vec{L}_{rod_i} | = r_i * m * v_i$

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

$| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}$

$| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}$

For our final angular momentum we have:

$| \vec{L}_{rod_f} | = r_f * m * v_f$

and the radius is 0.250 m and the mass is 2.00 kg

$| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f$

but, as the angular momentum is constant, this must be equal to the initial angular momentum

$7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f$

$v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}$

$v_f = 15 \frac{m}{s}$

8 0

3 years ago

A horse pulls on a wagon. The reaction force is the wagon pulls on the horse.<br><br> true or false?

Katen [24]

Answer:

True

Explanation:

the horse pull on the wagon but friction and the wight + gravity make the wagon pull on the horse ( newton's 2ed law)

6 0

2 years ago

Read 2 more answers

A box is hanging from two strings. String #1 pulls up and left, making an angle of 50° with the horizontal on the left, and stri

creativ13 [48]

Answer:

mb = 3.75 kg

Explanation:

System of forces in balance

ΣFx =0

ΣFy = 0

Forces acting on the box

T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left

T₂ = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.

Wb :Weightt of the box (vertical downward)

x-y T₁ and T₂ components

T₁x= T₁cos50°

T₁y= T₁sin50°

T₂x= 30*cos75° = 7.76 N

T₂y= 30*sin75° = 28.98 N

Calculation of the Wb

ΣFx = 0

T₂x-T₁x = 0

T₂x=T₁x

7.76 = T₁cos50°

T₁ = 7.76 /cos50° = 12.07 N

ΣFy = 0

T₂y+T₁y-Wb = 0

28.98 + 12.07(cos50°) = Wb

Wb = 36.74 N

Calculation of the mb ( mass of the box)

Wb = mb* g

g: acceleration due to gravity = 9.8 m/s²

mb = Wb/g

mb = 36.74 /9.8

mb = 3.75 kg

8 0

3 years ago

As part of the Antarctic Treaty System, all facilities on Antarctica are inspected. What is one reason there's a need for these

Alinara [238K]

Answer:

D. To ensure common resources are not overused

Explanation:

The article of the antarctic treaty VII provides that all the consultative parties have a right to designate the observers and to undertake the inspection as of the protocol 14 of that provides the protection of the environment and to safeguard the ecological system.

Since 1963, about 9 inspections have been conducted by the teams by visiting the stations, facilities, the vessels, and the protected areas across a wide area of East Antarctica, The most recent Antarctica expedition treaty inspection was done in Dec 2016.

5 0

2 years ago