Technician A says that hoods are designed with reinforcements to prevent folding during a collision. Technician B says that some
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A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.
Explanation:
From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).
To find the resistance of 260 ft (79.25 m) of size 4 AWG,
R= K * L/ A
K = 0.0214 ohm mm²/m
L = 79.25 m
A = 21.2 mm²
R = 0.0214 *
= 0.0214 * 3.738
= 0.0792 ohm.
Thus the resistance of uncoated copper wire is 0.0792 ohm
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
