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kumpel [21]
3 years ago
5

If "38 %" of the light passes through this combination of filters, what is the angle between the transmission axes of the filter

s
Physics
1 answer:
qaws [65]3 years ago
4 0

Answer:

<em>52°</em>

Explanation:

The initial intensity I_{0} = I

The final intensity I_{f} = 38% of I = 0.38

From the polarizing equation

I_{f} = I_{0} cos^{2}θ

substituting values, we have

0.38I = I cos^{2}θ

0.38 = cos^2θ

cosθ = \sqrt{0.38}

cosθ = 0.6164

θ = cos^{-1} 0.6164

θ = 51.9 ≅ <em>52°</em>

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A hydrogen atom that has lost its electron is moving east in aregion where the magnetic field is directed from south to north. I
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Answer:

a. up

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What acceleration would be achieved by a 5N thrust motor in a 0.30kg
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Which is a characteristic of a physical change?
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<span>The chemical bonds in the substance are unchanged.

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6 0
3 years ago
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An 800kg roller coaster starts from top of a 45m hill with a velocity of 4m/s. The car travels to the bottom, through a loop, an
aalyn [17]
Let's start with the total amount of energy available for the whole scenario:
Some kind of machine gave the coaster a bunch of potential energy by
dragging it up to the top of a 45m hill,and that's the energy is has to work with.

Potential energy = (M) (G) (H)  =  (800) (9.8) (45) = 352,800 joules

It was then given an extra kick ... enough to give it some kinetic energy, and
start it rolling at 4 m/s.

Kinetic energy = (1/2) (M) (V)² = (1/2) (800) (4)² = 6,400 joules

So the coaster starts out with (352,000 + 6,400) =<em> </em><u><em>359,200 joules</em></u><em> </em>of energy.

There's no friction, so it'll have <u>that same energy</u> at every point of the story.
=================================

Skip the loop for a moment, because the first question concerns the hill after
the loop.  We'll come back to it.

The coaster is traveling 10 m/sat the top of the next hill. Its kinetic energy is

(1/2) (M) (V)² = (400) (10)² = 40,000 joules.

Its potential energy at the top of the hill is (359,200 - 40,000) = 319,200.

PE = (M) (G) (H)

319,200 = (800) (9.8) (H)

H = (319,200) / (800 x 9.8) = <em>40.71 meters</em>
=================================

Now back to the loop:

You said that the loop is 22m high at the top. The PE up there is

PE = (M) (G) (H) = (800) (9.8) (22) = 172,480 joules

So the rest is now kinetic. KE = (359,200 - 172,480) = 186,720 joules.

KE = (1/2) (M) (V)² = 186,720

(400) (V)² = 186,720

V² = 186,720 / 400 = 466.8

V = √466.8 = <em>21.61 m/s</em>
===============================

Now it looks like there should be another question ... that's why they
bothered to tell you that the end is 4m off the ground. They must
want you to find the coaster's speed when it gets to the end.

At 4m off the ground,  PE = (M) (G) (H) = (800) (9.8) (4) = 31,360 joules.

The rest will be kinetic.  KE =  (359,200 - 31,360) = 327,840 joules

KE = (1/2) (M) (V)² = 327,840

400 V² = 327,840

V² = 327,840 / 400 = 819.6

V = √819.6 = <em>28.63 m/s</em> at the end
=======================================

If the official answers in class are a little bit different from these,
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I used '9.8' for gravity, but very often, they use '10' .

If the official answers in class are way way different from these,
then I made one or more big mistakes somewhere.  Sorry.
6 0
3 years ago
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