Answer:
Utilize the Standard Controller for Position_c and a Controller Extension to query for Review_c data.
Explanation:
Data types such as individual, pedigree, sample, and marker can be displayed depending on the type of information needed, and the database fields will be recorded based on their levels with the appropriate data type placed in each field. The relationship that will be gotten will depend on the auto-populated defaults which can changed to suit our requirements.
Answer: (C) 9.14 . 10⁻³ Ω
Explanation:
The resistance of a resistor, is proportional to his length and inversely proportional to his area, being the proportionality constant a property of the material, called resistivity.
The resistivity is defined as the inverse of the electrical conductivity, which depends on the number of charge carriers and the mobility of these carriers, which is different for each material.
So, we can calculate the resistance as follows:
R = 1/σ . L / A, where:
σ = electrical conductivity, l= length of the wire , A = wire cross-section (assumed circular).
Replacing by the values, we can calculate R as follows:
R = 1/6.1. 10⁷ (Ω.m) . 8.1 m. / π (0.0043)² m / 4 = 9.14 . 10⁻³ Ω
Answer:
a) Vout= 5V
b) Vout= 5V
c) Vbase= 0.6V
d) Vbase= 0.6V
Explanation:
Consider the circuit shown in attachment
a) When Vin is 0V, the base circuit is not turned, so
Ib=0 and Ic=∞ as transistor is not turned on so
Vout =5V
b) When Vin= 5 V,
Ib= (Vin-Vb)/Rb
Ib=(5-0.6)/1000= 0.0044A
Ic= 0.0044×10=0.044A
Vout= 5- 0.044×1000= not real value
Vout= Vce= 5V
c) voltage drop across Vbase= 0.6V
d) Vbase= 0.6V
In all the above cases, the transistor will not be turned on biasing base voltage and resistor values are very high compared to VCC which is 5V in the given circuit
Answer:
(a) The maximum volume flow rate for which the flow will be laminar is 0.0190 cubic meter per second
(b) The pressure drop required to deliver the maximum flow rate is 148962.96 Pascal
(c) The corresponding wall shear stress is 7600 Pascal
Explanation:
Reynolds number = 2299, density of water = 1000kg/m^3, diameter of needle = 0.27mm = 0.00027m, Length of needle = 50mm = 0.05m, viscosity of water = 0.00089kg/ms, area = 0.05m × 0.05m = 0.0025m^2, coefficient of friction = 64 ÷ Reynolds number = 64 ÷ 2299 = 0.028
Velocity = (Reynolds number × viscosity) ÷ (density × diameter) = (2299 × 0.00089) ÷ (1000 × 0.00027) = 2.046 ÷ 0.27 = 7.58m/s
(a) Maximum volume flow rate = velocity × area of needle = 7.58 × 0.0025 = 0.0190 cubic meter per second
(b) Pressure drop = ( coefficient of friction × length × density × velocity^2) ÷ (2 × diameter) = (0.028 × 0.05 × 1000 × 7.58^2) ÷ (2 × 0.00027) = 80.44 ÷ 0.00054 = 148962.96 Pascal
(c) Wall shear stress = (density × volume flow rate) ÷ area = (1000 × 0.0190) ÷ 0.0025 = 7600 Pascal
Answer
given,
Speed of vehicle = 65 mi/hr
= 65 x 1.4667 = 95.33 ft/s
e = 0.07 ft/ft
f is the lateral friction, f = 0.11
central angle,Δ = 38°
The PI station is
PI = 250 + 50
= 25050 ft
using super elevation formula
r = 1568 ft
As the road is two lane with width 12 ft
R = 1568 + 12/2
R = 1574 ft
Length of the curve
L = 1044 ft
Tangent of the curve calculation
T = 542 ft
The station PC and PT are
PC = PI - T
PC = 25050 - 542
= 24508 ft
= 245 + 8 ft
PT = PC + L
= 24508 + 1044
=25552
= 255 + 52 ft
the middle ordinate calculation
MO = 85.75 ft
degree of the curvature
D = 3.64°
