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xz_007 [3.2K]
4 years ago
12

Suppose that there are N=108 two-state systems, each with energy difference E=6 × 10-21 J between the two states. The environmen

t temperature is 300 K. The number of systems in their high energy states is M. Therefore, the free energy is a function of M: F(M).What is dU/dM?
Physics
1 answer:
kati45 [8]4 years ago
5 0

Answer:

The value of \dfrac{dU}{dM} is 6\times10^{-29}\ J/unit.

Explanation:

Given that,

Number N=10^{8}

Energy difference = 6\times10^{-21}\ J[/tex]

Temperature T =300 K

We need to calculate the value of \dfrac{dU}{dM}

We know that,

\dfrac{dU}{dM}=\dfrac{energy\ difference}{Change\ in\ number\ of\ system}

\dfrac{dU}{dM}=\dfrac{6\times10^{-21}}{10^{8}}

\dfrac{dU}{dM}=6\times10^{-29}\ J/unit

Hence, The value of \dfrac{dU}{dM} is 6\times10^{-29}\ J/unit.

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Evaporation (or another word to use is water vapor.)
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A nasa spacecraft measures the rate r of at which atmospheric pressure on mars decreases with altitude. the result at a certain
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Answer:4.21 \times 10^{-10} J/cm^4

1 kPa= 10^3 Pa

1 km=10^5 cm

1kPa/km=0.01 Pa/cm

1kPa/km=10^{-8} J/cm^4

\Rightarrow r= 0.0421 kPa/km= 0.0421 kPa/km \times \frac{10^{-8} J/cm^4}{1 kPa/km}= 0.0421 \times 10^{-8}J/cm^4=4.21 \times 10^{-10} J/cm^4

3 0
3 years ago
Read 2 more answers
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
Which statement correctly defines power?
Anna007 [38]

Answer:

Power=V\, I which corresponds to the second option shown: "voltage times amperage"

Explanation:

The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.

Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:

Power=\frac{V\,Q}{t} =V\,\frac{Q}{t} = V\, I

Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.

This corresponds to the second option shown in the question: "Voltage times amperage".

6 0
4 years ago
Identify each part of this chemical equation that describes the burning of methane and oxygen. B (blue box): D (number): E (purp
Allushta [10]

Answer:

The correct answer is -

A (the entire green box): Chemical Equation

B (the blue box): Reactants

C (the arrow): Reacts to Form

D (the number): Coefficient

E (the purple box): Products

Explanation:

The chemical reaction of burning methane and oxygen is as follows;

Here, the green part A is the chemical equation that includes various parts that are reactants B, methane, and oxygen, C is an arrow that indicates the formation of products.

2 is here coefficient that indicates the moles of the oxygen which forms carbon dioxide and water in box E is products

5 0
3 years ago
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