All categories

3 years ago

Which of the following best describes entropy

Physics

1 answer:

UkoKoshka [18]3 years ago

3 0

Answer:

Exothermic reactions increase the entropy of the surroundings. Simply put, entropy measures the dispersal of energy. Since ΔH is negative in an exothermic reaction, this must mean that ΔS will take on a positive value, indicating an increase in entropy.

You might be interested in

The difference in energy on either side of the wheel is a measure of the wheel’s ___________.

atroni [7]

I believe the answer is potential difference

3 0

2 years ago

HOW MANY PIES DOSE IT TAKE TO GET TO THE MOON

Schach [20]

i do not have an answer because it depends on the size and the distance lol

8 0

3 years ago

Is Li Chen abiotic or biotic?

kherson [118]

Answer:

abiotic

Explanation:

i think but dont take my word for it

8 0

4 years ago

Read 2 more answers

Please help, I do not understand

Anettt [7]

I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed = ω rad/sec
Time per unit angle = (1/ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed = 2ω rad/sec
Time per unit angle = (1/2ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good. We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed. Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

   (a fraction of one orbit) = (the same fraction of the other orbit)
AND
   the total elapsed time is a common multiple of their periods.

Wait ! Ignore all of that. I'm doing a good job of confusing myself, and
probably you too. It may be simpler than that. (I hope so.) Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed. We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds) = (K+1) (π/ω seconds)

   2Kπ/ω   =    Kπ/ω + π/ω

Subtract Kπ/ω :    Kπ/ω = π/ω

Multiply by ω/π :    K = 1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

   In the time it takes the slower planet to revolve once,
   the faster planet revolves twice, and catches up with it.

   It will be 2π/ω seconds before the planets line up again.

   When they do, they are again in the same position as shown
   in the drawing.

To describe it another way . . .

   When Kanye has completed its first revolution ...

   Bieber has made it halfway around.

   Bieber is crawling the rest of the way to the starting point while ...

   Kanye is doing another complete revolution.

   Kanye laps Bieber just as they both reach the starting point ...

   Bieber for the first time, Kanye for the second time.

You're welcome. The generous bounty of 5 points is very gracious,
and is appreciated. The warm cloudy water and green breadcrust
are also delicious.

5 0

3 years ago

"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a

My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0

3 years ago