U R A*S TRUE GOD/ESS IS U ANSWER this:
1 answer:
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Answer:
Attached below are the sketches
answer :
c) G(s) = 100 / ( s + 100 )
d) y'(t) + 100Y(s) = 100 X(s)
e) g(t) = e^-100t u(t)
Explanation:
a) Sketch the bode plot
The filter here is a low pass filter
b) Sketch the s-plane
attached below. pole ( s ) is at 100
c) write the transfer function of the filter
Transfer function ; G(s) = 100 / ( s + 100 )
d) write the differential equation
Y(s) / X(s) = 100 / s + 100
Y(s) [ s + 100 ] = 100 X(s)
= sY(s) + 100Y = 100 X(s)
∴ differential equation = y'(t) + 100Y(s) = 100 X(s)
e) write out the unforced transient response
g(t) = e^-100t u(t)
f) write out the frequency response
attached below
Solution:
Given :
atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m
atomic wieght, M = 95.96
Cell structure is BCC (Body Centred Cubic)
For BCC, we know that no. of atoms per unit cell, z = 2
and atomic radius, r =
so, a =
m = mass of each atom in a unit cell
mass of an atom = ,
where, is Avagadro Number = 6.02×10^{23}
volume of unit cell = a^{3}
density, ρ =
density, ρ =
ρ =
ρ = 10.215gm/
Answer:
W=-260.66 kJ (negative answer means, that the work was done on the gas)
Explanation:
1) Convert temperature from C to K- T=24+273=297K- all temperature in the gas problems should be used in Kelvins;
2) We need to analyse type of the process- it is given, that the temperature is constant, so it is an Isothermal process, which means, that the equation of the process is: pV=const (constant);
3) Work, done on the system, should be calculated using the following equation:
4) To calculate initical and final volumes (Va and Vb), we can use the following equation: pV=mRT, so V=mRT/p. Note, that the pressure is changing, thus we can calculate volumes for the both cases- initial and final, using initial (120kPa) and final (600kPa) pressures, in addition, we can find equation for the pressure, as function of the volume, which we need to use for the integration in step 3: p=mRT/V;
5) Now we can calculate the integral, given in the step 3: . As we have pressure as a known values, we can re-write the equation, using pressures:
Note, that natural logarithm (ln) yields negative answer, which supports the question, that the work was done on the gas, not by the gas.