Answer:
COP of heat pump=3.013
COP of cycle=1.124
Explanation
W = Q2 - Q1 ----- equation 1
W = work done
Q2 = final energy
Q1 = initial energy
A) calculate the COP of the heat pump
COP =Q2/W
from equation 1
Q2 = Q1 + W = 15 + 7.45 = 22.45 KW
therefore COP =22.45/7.45 = 3.013
B) COP when cycle is reversed
COP = Q1/W
from equation 1
Q1 + W = Q2 ------ equation 2
Q2 = 15 Btu/s = 15 * 1.055 = 15.825 KW therefore from equation 2
Q1 = 8.375 KW
COP =8.375/7.45 = 1.124
Answer: 1766.667 Ω = 1.767kΩ
Explanation:
V=iR
where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).
3mA = 0.003 A
Rearranging the equation, we get
R=V/i
Now we are solving for resistance. Plug in 0.003 A and 5.3 V.
R = 5.3 / 0.003
= 1766.6667 Ω
= 1.7666667 kΩ
The 6s are repeating so round off to whichever value you need for exactness.
Answer:
conduction occurs when a substance is heated
Explanation:
Answer:
The given grammar is :
S = T V ;
V = C X
X = , V | ε
T = float | double
C = z | w
1.
Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.
From the given grammar,
Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.
No other variables generate variable X or ε.
So, only variable X is nullable.
2.
First of nullable variable X is First (X ) = , and ε (epsilon).
L.H.S.
The first of other varibles are :
First (S) = {float, double }
First (T) = {float, double }
First (V) = {z, w}
First (C) = {z, w}
R.H.S.
First (T V ; ) = {float, double }
First ( C X ) = {z, w}
First (, V) = ,
First ( ε ) = ε
First (float) = float
First (double) = double
First (z) = z
First (w) = w
3.
Follow of nullable variable X is Follow (V).
Follow (S) = $
Follow (T) = {z, w}
Follow (V) = ;
Follow (X) = Follow (V) = ;
Follow (C) = , and ;
Explanation: