Fastest
(Known as the fast lane)
Answer:a
a) Vo/Vi = - 3.4
b) Vo/Vi = - 14.8
c) Vo/Vi = - 1000
Explanation:
a)
R1 = 17kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0
sin we know Va≈Vb=0
so
-Vi/5kΩ + -Vo/17kΩ = 0
Vo/Vi = - 17k/5k
Vo/Vi = -3.4
║Vo/Vi ║ = 3.4 ( negative sign phase inversion)
b)
R2 = 74kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
so
(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0
-Vi/5kΩ + -Vo/74kΩ = 0
Vo/Vi = - 74kΩ/5kΩ
Vo/Vi = - 14.8
║Vo/Vi ║ = 14.8 ( negative sign phase inversion)
c)
Also for ideal op-amp
Va≈Vb=0 so Va=0
Now for position 3 we apply nodal analysis we got at position 1
(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0 ( 5MΩ = 5000kΩ )
so
-Vi/5kΩ + -Vo/5000kΩ = 0
Vo/Vi = - 5000kΩ/5kΩ
Vo/Vi = - 1000
║Vo/Vi ║ = 1000 ( negative sign phase inversion)
Answer:
The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.
Explanation:
Diameter of pipe = 2 in = 0.0508 m
Steam temperature 
= 300 F = 422.04 K
Duct temperature 
= 70 F = 294.26 K
Emmisivity of surface 1 = 0.79
Emmisivity of surface 2 = 0.276
Net emmisivity of both surfaces ∈ = 0.25
Stefan volazman constant 
= 5.67 × 
Heat transfer per foot length is given by
Q = ∈ A ( 
) ------ (1)
Put all the values in equation (1) , we get
Q = 0.25 × 5.67 × × 3.14 × 0.0508 × 1 × ( 
)
Q = 54.78 Watt per foot.
This is the value of heat transferred watt per foot length.
Answer:
See explanation
Explanation:
The magnetic force is
F = qvB sin θ
We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields
F
=
(
20
×
10
−
9
C
)
(
10
m/s
)
(
5
×
10
−
5
T
)
=
1
×
10
−
11
(
C
⋅
m/s
)
(
N
C
⋅
m/s
)
=
1
×
10
−
11
N
Answer:
I always thought it was so that the older wire could not have a problem and have another electrician must come back and fix it.
Explanation:
