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Dahasolnce [82]

3 years ago

15

Ethan made a diagram to compare examples of the first and second laws of thermodynamics. What belongs in the areas marked X and

Y?

2 answers:

Thepotemich [5.8K]3 years ago

3 0

<u>First law of thermodynamics:</u>

It states that <em>"Energy neither be created nor it can be destroyed". </em>simply it converts one form of energy into another form.

It is also known as<em> "law of conservation of energy"</em>

<u>Limitations of First law</u>

It doesn't provide a clear idea about the direction of transfer of heat.
It doesn't provide the information that how much heat energy converted inti work.
Its not given any practical applications.

<u>II law of thermodynamics:</u>

It states that <em>"the total entropy of the system can never decrease over time"</em>

It is strongly proved by two laws, they are

<em>1. Kelvin-plank statement:</em>

He stated that "any engine does not give 100% efficiency". It violates the Perpetual motion of machine II kind<em>(PMM-II).</em>

<em>2. Classius statement: </em>

<em> </em><em> It states that "Heat always flows from high temperature body to low temperature body, without aid of external energy". </em>

<em> Also it stated that " Heat can also be transferred from low temperature body to high temperature body, by the aid of an external energy".</em>

<em>Applications of II law: </em>

<em>Refrigeration &Air conditioning, Heat transfer, I.C. engines, etc.</em>

KatRina [158]3 years ago

3 0

Answer:

The answer is D.

Explanation:

I just took the quiz.

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It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon

trapecia [35]

Answer: $8.76\times 10^{-3} min^{-1}$

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

$y=1-e^{-kt^n}$

where y=fraction Transformed

k=constant

t=time

$0.3=1-e^{-k(100)^5}$

$e^{-k(100)^5} =0.7$

Taking log both sides

$-k\cdot (10^{10}=\ln 0.7$

$k=3.566\times 10^{-11}$

At this Point we want to compute $t_{0.5}\ i.e.\ y=0.5$

$0.5=1-e^{-kt^n}$

$0.5=e^{-kt^n}$

$0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}$

taking log both sides

$\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5$

$t^5=1.943\times 10^{10}$

$t=114.2 min$

Rate of Re crystallization at this temperature

$t^{-1}=8.76\times 10^{-3} min^{-1}$

3 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

Power of an electric motor is 1 h.p. what does it mean ?

const2013 [10]

Answer: HP = Horse Power.

Explanation: it is the unit given to tell the motor's particular power and 1hp = 746 watts.

8 0

2 years ago

Please help :)<br><br><br> Microwaves are transmitted by ....... ?

Dima020 [189]

Microwaves are transmitted by Radio Waves.

4 0

3 years ago

A police car chases a speeder along a straight road towards a cliff both vehicles move at 160km/h the siren on the police car pr

natta225 [31]

Answer:

f ’= 97.0 Hz

Explanation:

This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer

in this case the source is the police cases that go to vs = 160 km / h

and the observer is vo = 120 km / h

the relationship of the doppler effect is

f ’= f₀ (v + v₀ / v- $v_{s}$ )

let's reduce the magnitude to the SI system

v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s

v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s

we substitute in the equation of the Doppler effect

f ‘= 100 (330+ 33.33 / 330-44.44)

f ’= 97.0 Hz

4 0

3 years ago