Answer:
Explanation:
Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)
fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)
vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)
BFFS = 50+5, BFFS =55 (given) fLW= 6.6
TLC=6+3=9 fLC= 0.65
fM= 0.0
fA= 1.0
FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)
Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)
After: fA= 3.0
FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)
Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln
Answer:
The function of empirical evidence is information that verifies the "TRUTH" (reality) or falsity of a claim. In the empiricist view, you can only claim to have knowledge based on "empirical evidence"
I am not extremely sure that this is 100% correct because this is only based on my own opinion and understanding. "Empirical Evidence" to me is knowledge toward the reality of the world, finding the truth within the unknown discoveries.
I can't say that this will really help you but I hope that it'll provide you some kind of idea...?!
Explanation:
Answer:
(a) 1.003 MHz
(b) increase the signal-to-noise ratio
Explanation:
The relationship between channel capacity, bandwidth, and signal-to-noise ratio was described by Claude Shannon in the 1940s, based on previous work by Harry Nyquist and Ralph Hartley. The relationship is given by the formula ...
C = Blog₂(1 +S/N)
where C is the channel capacity (bits per second), B is the bandwidth (Hz), and S/N is the signal to noise power ratio.
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<h3>(a)</h3>
You want to find B for C=10^7 and 10·log(S/N) = 30. Solving for B, we find ...
B = C/log₂(1 +S/N)
B = 10^7/log₂(1 +10^(30/10)) = 10^7/log₂(1001) ≈ 1.003 MHz
The allocated bandwidth must be at least 1.003 MHz.
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<h3>(b)</h3>
Shannon's relation tells us the alternative to increasing bandwidth is increasing signal-to-noise ratio. If "sufficient" bandwidth is not available, the signal-to-noise ratio must be increased.
Answer:
look below at illustrated diagrams and solution for Mz=1.13kN
Answer:
so heat loss = 4312 W
cost of heat loss daily is $8.28 per day
Explanation:
given data
slab length L = 11 m
slab wide W = 8 m
thickness t = 0.20 m
temperature top T1 = 17°C
temperature bottom T2 = 10°C
thermal conductivity k = 1.4 W/m-K
efficiency ηf = 0.90
priced Cg = $0.02 / MJ
to find out
rate of heat loss and daily cost of the heat loss
solution
we calculate here heat loss by heat transfer
so apply here formula that is
q = (thermal conductivity × area × temperature difference) / thickness
put here all these value we get heat loss
q =
q =
q = 4312 W
so heat loss = 4312 W
and
cost of the heat loss is express as
cost of heat loss =
put here all these value
cost of heat loss is = × 24 hr/day × 3600 s/hr
cost of heat loss = 8.279
so cost of heat loss daily is $8.28 per day