The outer surface temperature of a glass filled with iced water may drop below the dew-point temperature of the surrounding air,
1 answer:
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Answer:
The resistance is 24.9 Ω
Explanation:
The resistivity is equal to:
The area is:
A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²
If NA is greater, then, the term 1/NA can be neglected, thus the equation:
Where
V = 0.44 V
E = 11.68*8.85x10¹⁴ f/cm
The length is:
L = 10 - 0.335 = 9.665 um
The resistance is:
Answer: maximum length of the nanowire is 510 nm
Explanation:
From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K
Thermal conductivity of silicon carbide k = 30 W/m.K
Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m
lets consider the equation for the value of m
m = ( (hP/kAc)^1/2 ) = ( (4h/kD)^1/2 )
m = ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04
now lets find the value of h/mk
h/mk = 10⁵ / ( 942809.04 × 30) = 0.00353
lets consider the value θ/θb by using the equation
θ/θb = (T - T∞) / (T - T∞)
θ/θb = (3000 - 8000) / (2400 - 8000)
= 0.893
the temperature distribution at steady-state is expressed as;
θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)] / [cosh mL+ (h/mk) sinh mL]
θ/θb = [ 1 ] / [cosh mL+ (h/mk) sinh mL]
so we substitute
0.893 = [ 1 ] / [cosh (942809.04 × L) + (0.00353) sinh (942809.04 × L)]
L = 510 × 10⁻⁹m
L = 510 nm
therefore maximum length of the nanowire is 510 nm
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)
Now we can use the equation that defines the percentage of increase, in this case for speed
Now we use the equation obtained in the previous step, and replace values
the percent increase in the velocity of air is 25.65%