Answer:
View Image
Explanation:
You didn't provide me a picture of the opamp.
I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...
- no current will go in the inverting(-) and noninverting(+) side of the opamp
- V₊ = V₋ , so whatever voltage is at the noninverting side will also be the voltage at the inverting side
Since no current is going into the + and - side of the opamp, then
i₁ = i₂
Since V₊ is connected to ground (0V) then V₋ must also be 0V.
V₊ = V₋ = 0
Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.
You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.
Answer:
final volume V2 = 0.71136 m³
work done in process W = -291.24 kJ
heat transfer Q = 164 kJ
Explanation:
given data
mass = 1.5 kg
pressure p1 = 200 kPa
temperature t1 = 150°C
final pressure p2 = 600 kPa
final temperature t2 = 350°C
solution
we will use here superheated water table that is
for pressure 200 kPa and 150°C temperature
v1 = 0.95964 m³/kg
u1 = 2576.87 kJ/kg
and
for pressure 600 kPa and 350°C temperature
v2 = 0.47424 m³/kg
u2 = 2881.12 kJ/kg
so v1 is express as
V1 = v1 × m ............................1
V1 = 0.95964 × 1.5
V1 = 1.43946 m³
and
V2 = v2 × m ............................2
V2 = 0.47424 × 1.5
final volume V2 = 0.71136 m³
and
W = P(avg) × dV .............................3
P(avg) = 
= 
= 400 × 10³
put here value
W = 400 × 10³ × (0.71136 - 1.43946 )
work done in process W = -291.24 kJ
and
heat transfer is
Q = m × (u2 - u1) + W .............................4
Q = 1.5 × (2881.12 - 2576.87) + 292.24
heat transfer Q = 164 kJ
Answer & Explanation:
function Temprature
NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];
DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];
%AVERAGE CALCULATION AND ROUND TO NEAREST INT
avgNYC=round(mean(NYC));
avgDEN=round(mean(DEN));
fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);
fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);
%part B
count=1;
NNYC=0;
NDEN=0;
while count<=length(NYC)
if NYC(count)>avgNYC
NNYC=NNYC+1;
end
if DEN(count)>avgDEN
NDEN=NDEN+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);
fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);
%part C
count=1;
highDen=0;
while count<=length(NYC)
if NYC(count)>DEN(count)
highDen=highDen+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);
end
%output
check the attachment for additional Information
Answer:
it is not possible to place the wires in the condui
Explanation:
given data
total area = 2.04 square inches
wires total area = 0.93 square inches
maximum fill conduit = 40%
to find out
Can it is possible place wire in conduit conduit
solution
we know maximum fill is 40%
so here first we get total area of conduit that will be
total area of conduit = 40% × 2.04
total area of conduit = 0.816 square inches
but this area is less than required area of wire that is 0.93 square inches
so we can say it is not possible to place the wires in the conduit
Answer:
230.51 m
Explanation:
Pb = 695 mmHg
Pt = 675 mmHg
Pb - Pt = 20 mmHg
Calculate dP:
dP = p * g * H = (13600)*(9.81)*(20/1000) = 2668.320 Pa
Calculate Height of building as dP is same for any medium of liquid
dP = p*g*H = 2668.320
H = 2668.32 / (1.18 * 9.81) = 230.51 m
