Answer:
See the attachment
Verification shown in explanation
Explanation:
wc= 1/(RC)
The value of R and C shown in the attachment can be calculated as follows:
R=100 ohms
1000= 1/(100×C)
C= 10 microfarads
Gain= R/√(R²+Xc²)
at f=500, wc= 3142.59 rad/s
Xc= 1/(2πfC)
Xc=31.83
Gain= 100/√(100²+31.83²)
Gain= 0.953
at fc= 100, wc= 628.32 rad/s
Xc= 159.15
gain= 100/√(100²+ 159.15²)
gain= 0.532
As see form calculations that gain at low frequencies is smaller and gain at very high frequencies is near to 1. This means that at high frequencies Vin≈ Vout
Hence this filter allows high frequencies to pass but blocks low frequencies by lowering the gai
Answer:
Compute the number of gold atoms per cubic centimeter = 9.052 x 10^21 atoms/cm3
Explanation:
The step by step and appropriate substitution is as shown in the attachment.
From number of moles = Concentration x volume
number of moles = number of particles/ Avogadro's number
Volume = mass/density, the appropriate derivation to get the number of moles of atoms
Answer:
maximum allowable electrical power=4.51W/m
critical radius of the insulation=13mm
Explanation:
Hello!
To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)
Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and conduction as shown in the attached image
to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient
<u>Explanation:</u>
% Gravel = 100 - 72% = 28%
% Sand = 100 - 28 - 15 = 57%
% Fires = 100 - 0 - 28 - 57 = 15%
The soil is poorly graded soil.
