Answer:
hello your question is incomplete attached below is the complete question
answer : attached below
Explanation:
let ; x(t) be a real value signal for x ( jw ) = 0 , |w| > 200
g(t) = x ( t ) sin ( 2000 

next we apply Fourier transform
attached below is the remaining part of the solution
Answer:
± 0.003 ft
Explanation:
Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.
Let L' = our distance and L = our tape measure
So, L' = 100L
Now by error determination ΔL' = 100ΔL
Now ΔL' = ± 0.30 ft
ΔL = ΔL'/100
= ± 0.30 ft/100
= ± 0.003 ft
So, the maxim error per tape is ± 0.003 ft
Answer:
The theoretical maximum specific gravity at 6.5% binder content is 2.44.
Explanation:
Given the specific gravity at 5.0 % binder content 2.495
Therefore
95 % mix + 5 % binder gives S.G. = 2.495
Where the binder is S.G. = 1, Therefore
Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495
(95 +5)/(Vx +5) = 2.495
2.495 × (Vx + 5) = 100
Vx =35.08 to 95
Or density of mix = Mx/Vx = 95/35.08 = 2.7081
Therefore when we have 6.5 % binder content, we get
Per 100 mass unit
93.5 Mass unit of Mx has a volume of
Mass/Density = 93.5/2.7081 = 34.526 volume units
Therefore we have
At 6.5 % binder content.
(100 mass unit)/(34.526 + 6.5) = 2.44
The theoretical maximum specific gravity at 6.5% binder content = 2.44.
In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.
With the given data we can proceed to calculate the compression stress:



Through Goodman's equations the combined effort by fatigue and compression is expressed as:

Where,
Fatigue limit for comined alternating and mean stress
Fatigue Limit
Mean stress (due to static load)
Ultimate tensile stress
Security Factor
We can replace the values and assume a security factor of 1, then

Re-arrenge for 

We know that the stress is representing as,

Then,
Where
=Max Moment
I= Intertia
The inertia for this object is

Then replacing and re-arrenge for 



Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm
Answer:
The mass flow rate of steam m=5.4 Kg/s
Explanation:
Given:
At the inlet of turbine P=10 MPa ,T=500 C
AT the exit of turbine P=10 KPa ,x=0.9
Required power=5 MW
From steam table
<u> At 10 MPa and 500 C:</u>
h=3374 KJ/Kg ,s=6.59 KJ/Kg-K (Super heated steam table)
<u>At 10 KPa:</u>
=2675.1 KJ/Kg,
=417.51 KJ/Kg
= 7.3 KJ/Kg-K ,
=1.3 KJ/Kg-K
So enthalpy of steam at the exit of turbine
h= 
Now by putting the values
h= 417.51+0.9(2675.1- 417.51) KJ/Kg
h=2449.34 KJ/Kg
Lets take m is the mass flow rate of steam
So 
m=5.4 Kg/s
So the mass flow rate of steam m=5.4 Kg/s