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4 years ago

Process of changing a liquid into a gas

Physics

2 answers:

Komok [63]4 years ago

7 0

The process of changing a liquid to a gas is called evaporation.

lions [1.4K]4 years ago

3 0

The process is called evaporation. Liquid turns into gas.

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Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

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. A long 10-cm-diameter steam pipe whose external surface temperature is 110oC passes through some open area that is not protect

Nata [24]

Answer:

$Nu = 30.311$

Explanation:

Let consider that pipe is a horizontal cylinder. The Nusselt number is equal to:

$Nu = \left\{0.6+\frac{0.387\cdot Ra_{D}^{\frac{1}{6} }}{[1+\left(\frac{0.559}{Pr} \right)^{\frac{9}{16}} ]^{\frac{8}{27} }} \right\}^{2}$ , for $Ra_{D} \le 10^{12}$ .

Where $Ra_{D}$ is the Rayleigh number associated with the cylinder.

The Rayleigh number is:

$Ra_{D} = \frac{g\cdot \beta\cdot (T_{pipe}-T_{air})\cdot D^{3}}{\nu^{2}}\cdot Pr$

By assuming that air behaves ideally, the coefficient of volume expansion is:

$\beta = \frac{1}{T}$

$\beta = \frac{1}{283.15\,K}$

$\beta = 3.532\times 10^{-3}\,\frac{1}{K}$

The cinematic and dynamic viscosities, thermal conductivity and isobaric specific heat of air at 10 °C and 1 atm are:

$\nu = 1.426\times 10^{-5}\,\frac{m^{2}}{s}$

$\mu = 1.778\times 10^{-5}\,\frac{kg}{m\cdot s}$

$k = 0.02439\,\frac{W}{m\cdot ^{\textdegree}C}$

$c_{p} = 1006\,\frac{J}{kg\cdot ^{\textdegree}C}$

The Prandtl number is:

$Pr = \frac{\mu\cdot c_{p}}{k}$

$Pr = \frac{(1.778\times 10^{-5}\,\frac{kg}{m\cdot s} )\cdot (1006\,\frac{J}{kg\cdot ^{\textdegree}C} )}{0.02439\,\frac{W}{m\cdot ^{\textdegree}C} }$

$Pr = 0.733$

Likewise, the Rayleigh number is:

$Ra_{D} = \frac{(9.807\,\frac{m}{s^{2}} )\cdot (3.532\times 10^{-3}\,\frac{1}{K} )\cdot (110^{\textdegree}C-10^{\textdegree}C)\cdot (0.1\,m)^{3}}{(1.426\times 10^{-5}\,\frac{m^{2}}{s})^{2} }\cdot (0.733)$

$Ra_{D} = 12.486\times 10^{6}$

Finally, the Nusselt number is:

$Nu = \left\{0.6+\frac{0.387\cdot (12.486\times 10^{6})^{\frac{1}{6} }}{\left[1 + \left(\frac{0.559}{0.733}\right)^{\frac{9}{16} }\right]^{\frac{8}{27} }} \right\}^{2}$