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3 years ago

5

The outer surface temperature of a glass filled with iced water may drop below the dew-point temperature of the surrounding air,

causing the moisture in the vicinity of the glass to condense. After a while, the condensate may start dripping down because of gravity.
1. True
2. False

1 answer:

user100 [1]3 years ago

3 0

Answer: True

Explanation:

The outer surface temperature of a glass filled with iced water will drop below dew-point temperature of the surrounding air, thereby resulting into the moisture in the vicinity of the glass to condense.

After some time, the condensate may start dripping down because of gravity. This is due to the fact that water will start condensing on glass surface below due point temperature. The adhesion force that water has with glass is low.

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When a technician is ohming out a load in a circuit, what reading should the technician get if the load is good

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When a technician is ohming out a load in a circuit, the reading that the technician should get if the load is good is some numbers.

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2 years ago

Saturated steam coming off the turbine of a steam power plant at 30 °C condenses on the outside of a 3-cm-outer-diameter, 35-m-l

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Answer:

The answer 30.07J/sec

Explanation:

GIVEN DATA:

temperature of steam = 40 degree celcius

mass flow rate = 63 kg/h

from saturated water tables,

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3 years ago

Technician A says that in a circuit that has 3 Ohms of resistance and operates on 12 volts, the amperage must be 4 amps. Technic

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3 years ago

The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in t

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Answer:

Explanation:

$r_2=$ ∞

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$q=2\pi kD.$ ΔT--------(1)

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$N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)$

By equation (1) and (2)

$2\pi kD.$ ΔT=h.4 $\pi r_1^2$ ΔT

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From equation (3) and (4)

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3 years ago

A particular NMOS device has parameters VT N = 0.6 V, L = 0.8µm, tox = 200 Å, and µn = 600 cm2 /V–s. A drain current of ID = 1.2

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Answer:

$W= 3.22 \mu m$

Explanation:

the transistor In saturation drain current region is given by:

$i_D}=K_a(V_{GS}-V_{IN})^2$

Making $K_a$ the subject of the formula; we have:

$K_a=\frac {i_D} {(V_{GS} - V_{IN})^2}$

where;

$i_D = 1.2m$

$V_{GS}= 3.0V$

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$K_a=\frac {1.2m} {(3.0 - 0.6)^2}$

$K_a = 208.3 \mu A/V^2$

Also;

$k'_n}=\frac{\mu n (\frac{cm^2}{V-s} ) \epsilon _{ox}(\frac{F}{cm} ) }{t_{ox}(cm)}$

where:

$\mu n (\frac{cm^2}{V-s} ) = 600$

$\epsilon _{ox}=3.9*8.85*10^{-14}$

${t_{ox}(cm)=200*10^{-8}$

substituting our values; we have:

$k'_n}=\frac{(600)(3.988.85*10^{-14})}{(200*10^{-8})}$

$k'_n}=103.545 \mu A/V^2$

Finally, the width can be calculated by using the formula:

$W= \frac{2LK_n}{k'n}$

where;

L = $0.8 \mu m$

$W= \frac{2*0.8 \mu m *208.3 \mu}{103.545 \mu}$

$W= 3.22 \mu m$

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4 years ago