How much force is required to drag a 90 lb. box up this "frictionless" inclined plane? 109 lb. 10 lb. 81 lb. 9 lb.

Some unpolarized light has an intensity of 1375 W/m2 before passing through three polarizing filters. The transmission axis of t

chubhunter [2.5K]

Answer:

intensity of the light that emerges from the three filters is 560.80 W/m²

Explanation:

Given data

intensity I = 1375 W/m2

angle 1 = 31.0°

angle 2 = 41.0°

to find out

intensity of the light that emerges from the three filters

solution

we know intensity of light pass 1st polarize = I/2 = 1375 / 2 = 687.5 W/m2

so intensity after 2nd polarize pass = I 1st cos²(θ)

I 2nd = 687.5 cos²(31) = 687.5 ( 0.836754) = 575.27 W/m2

and

intensity after 3rd polarize pass = I 2nd cos²(θ)

I 3rd = 575.27 cos²(41) = 575.27 (0.974839) = 560.80 W/m2

so that intensity of the light that emerges from the three filters is 560.80 W/m²

4 0

4 years ago

If white light falls on a piece of white paper in a dark room, the paper appears white to the eye. This is because the paper ref

Sindrei [870]

It would appear Red because as stated above, the paper reflects all the colors of light that fall on it.

6 0

3 years ago

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Which planet has a rocky mantle, iron core and very dense atmosphere that is 96.5% poisonous carbon dioxide and 3.5% nitrogen?

pochemuha

"Venus" has all that characteristics. It has <span>a rocky mantle, iron core and very dense atmosphere that is 96.5% poisonous carbon dioxide and 3.5% nitrogen

In short, Your Answer would be "Venus"

Hope this helps!</span>

3 0

3 years ago

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How does an objects size affect the gravity exerted by that object

egoroff_w [7]

Answer:

The size of an object is directly proportional to the gravity

Explanation:

The size of an object has significant impact on the gravity exerted by such a body.

The more massive a body is, the larger the gravity it exerts.

The reason for this is because of the newton's law of universal gravitation.

It states that "the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".
As such, gravity is directly proportional to mass

4 0

3 years ago

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 55.0°-angle with the horizontal. (a

Nikitich [7]

Answer:

Check attachment for the free body diagram

Explanation:

Given that,

Ladder length L = 14m

Weight of ladder W= 490N

The weight will act at the midpoint

i.e at 14/2 = 7m, L1 = 7m

The ladder makes an angle of 55° with the horizontal. θ=55°

Weight of firefighter Wf =810N

The firefighter is at 3.9m from the horizontal ground, L2 =3.9

The wall exerts a force on the ladder, let It be Nw

The ground exerts a force on the ladder, let it be Ng

The let Ff be the frictional force that opposes motion.

a. We want to find the horizontal and vertical force the ground exerted on the ladder i.e Ng and Ff.

Using Newton second law

ΣFy= m•ay

ay=0, since the body is not accelerating

ΣFy = 0

Ng — Wf —W = 0

Ng = Wf + W

Ng = 490 + 810

Ng = 1300 N.

Also,

ΣFx= m•ax

ax=0, since the body is not accelerating

ΣFx = 0

Ff — Nw = 0

Ff = Nw

Now,

Let take moment about point A(ground), but note before we take moment, the forces must be perpendicular to the ladder.

applying condition of equilibrium of moment

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•3.9 + (490Cos55)•7 = (NwSin55)•14

1811.93 + 1967.37 = 11.47Nw

3779.3 = 11.47Nw

Then, Nw = 3779.3/11.47

Nw= 329.49N

Since Nw = Ff

Then Ff = 329.49N

So the required reaction exerted by the ground on the ladder are

Ng = 1300 N

Ff = 329.49 N

b. Now the firefighter is a distance of 9.4m from the horizontal and the ladder is about to slip

So we need to calculate the coefficient of static friction μs

Check attachment for new diagram,

So calculating for Nw again, since the firefighter have new position

Now the firefighter is at 9.4m from the ground. Therefore, L2 = 9.4m

applying condition of equilibrium of moment

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•9.4 + (490Cos55)•7 = (NwSin55)•14

4367.21 + 1967.37 = 11.47Nw

6334.58 = 11.47Nw

Then, Nw = 6334.58/11.47

Nw= 552.274N

Since Nw = Ff

Then Ff = 552.274N

Then, using frictional law

Ff = μs•Ng

The Ng doesn't change

Then, 552.274 = 1300μs

μs = 552.274/1300

μs = 0.42

c. Now, we want to know the maximum distance of the firefighter if the coefficient of static friction is reduce by half

Then, μs = 0.42/2

μs = 0.21

The assume the firefighter is at L2 from the horizontal.

Then, the frictional force is

Ff = μsNg

Ff = 0.21 × 1300

Ff = 273N

Then, Nw = Ff = 273N

Taking moment about point A

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•L2 + (490Cos55)•7 = (273Sin55)•14

464.6•L2 + 1967.37 = 3130.8

464.6•L2 = 3130.8—1967.37

464.6L2 = 1163.43

L2 = 1163.43/464.6

L2 = 2.5m

The maximum distance before the ladder begin to slip is 2.5m.

3 0

3 years ago

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