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2 years ago

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1. Phát biểu và viết biểu thức nguyên lý I nhiệt động học, giải thích các ký hiệu.

1 answer:

Alisiya [41]2 years ago

4 0

Answer:

hmm mm which language is this

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The equation that is used to solve second law problems is # F= ma.

maw [93]

F= Force
M=Mass
A= acceleration
F=N
Mass= in grams or kilo grams (mostly kg)
A= m/s

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3 years ago

Which change would decrease the total current, I, flowing through this circuit?

iren [92.7K]

Answer:

A. Increasing the voltage of the battery

Explanation:

The relationship between voltage, V, current, I and resistance, R, is given as follows;

V = I × R

∴ I = V/R

From the above relationship, the current flowing in the circuit is directly proportional to the voltage of the battery, and inversely proportional to the resistance, 'R', of the circuit

Therefore, increasing the voltage, 'V', of the battery, increases the total current, 'I', flowing in the circuit.

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3 years ago

Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are

Juliette [100K]

Answer: $Q=248.011 W$

Explanation:

Given

Temperature of Room $T_{\infty }=18^{\circ}\approx 291 K$

Area of Person $A=1.7 m^2$

Temperature of skin $T=32^{\circ}\approx 305 K$

Heat transfer coefficient $h=5 W/m^2.k$

Emissivity of the skin and clothes $\epsilon =0.9$

$\Delta T=32-18=14^{\circ}$

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection

Heat transfer due radiation $Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )$

where $\sigma =stefan-boltzman\ constant$

$Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)$

$Q_1=129.01 W$

Heat Transfer due to convection is given by

$Q_2=hA(\Delta T)$

$Q_2=5\times 1.7\times 14=119 W$

$Q=Q_1+Q_2$

$Q=129.01+119=248.011 W$

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4 years ago

One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at

Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature $(T_i)=300 K$

$P_i=10 atm$

$P_f=2 atm$

Work done in iso-thermal process $=P_iV_iln\frac{P_i}{P_f}$

$P_i$ =initial pressure

$P_f$ =Final Pressure

$W=10\times 2.463\times ln\frac{10}{2}=39.64 J$

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is $=P\Delta V$

$V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L$

$V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L$

$\Delta W=1\times (12.315-2.463)=9.852 J$

$\Delta q=\Delta W=9.852 J$

$\Delta U=0$

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4 years ago

A 70 Newton dog ran across the yard. What do we know about the dog?

Amiraneli [1.4K]

The dogs acceleration

5 0

3 years ago