Answer:
The AGC circuit operates with an input voltage range of 60 dB (5 mV p-p to 5 V p-p), with a fixed output voltage of 250 mV p-p.
Explanation:
Answer:
%Reduction in area = 73.41%
%Reduction in elongation = 42.20%
Explanation:
Given
Original diameter = 12.8 mm
Gauge length = 50.80mm
Diameter at the point of fracture = 6.60 mm (0.260 in.)
Fractured gauge length = 72.14 mm.
%Reduction in Area is given as:
((do/2)² - (d1/2)²)/(do/2)²
Calculating percent reduction in area
do = 12.8mm, d1 = 6.6mm
So,
%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²
%RA = 0.734130859375
%RA = 73.41%
Calculating percent reduction in elongation
%Reduction in elongation is given as:
((do) - (d1))/(d1)
do = 72.14mm, d1 = 50.80mm
So,
%RA = ((72.24) - (50.80))/(50.80)
%RA = 0.422047244094488
%RA = 42.20%
Answer:
power developed by the turbine = 6927.415 kW
Explanation:
given data
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2431.7 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 11.95 kg/s
solution
we apply here thermodynamic equation that
energy equation that is

put here value with
turbine is insulated so q = 0
so here

solve we get
w = 579700 J/kg = 579.7 kJ/kg
and
W = mass flow rate × w
W = 11.95 × 579.7
W = 6927.415 kW
power developed by the turbine = 6927.415 kW
Answer:
Final length of the rod = 13.90 in
Explanation:
Cross Sectional Area of the polythene rod, A = 0.04 in²
Original length of the polythene rod, l = 10 inches
Tensile modulus for the polymer, E = 25,000 psi
Viscosity, 
Weight = 358 lbs - f
time, t = 1 hr = 3600 sec
Stress is given by:

Based on Maxwell's equation, the strain is given by:

Strain = Extension/(original Length)
0.39022 = Extension/10
Extension = 0.39022 * 10
Extension = 3.9022 in
Extension = Final length - Original length
3.9022 = Final length - 10
Final length = 10 + 3.9022
Final length = 13.9022 in
Final length = 13.90 in
Answer:
Maximum number that can be represented by 13 bits is 8192 Instructions
Explanation:
number of instructions = 1000
number of bits = log(1000) x number of register
= 6 bits
Since the complete instruction must have 32 bits, then
remaining number of bits = 32 - 6 = 236
number of registers in instruction = 2
number of bits per register = 26/2 = 13
Maximum number that can be represented by 13 bits = 
= 2¹³ = 8192