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:
F=1.47 KN
Explanation:
Given that
Diameter of plate = 25 cm
Height of pool h = 3 m
We know that force can be given as
F= P x A
P=ρ x g x h
Now by putting the values
P=1000 x 10 x 3
P= 30 KPa
![A=\dfrac{\pi}{4}\times 0.25^2\ m^2](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Ctimes%200.25%5E2%5C%20m%5E2)
![A=0.049\ m^2](https://tex.z-dn.net/?f=A%3D0.049%5C%20m%5E2)
F= 30 x 0.049 KN
F=1.47 KN
So the force on the plate will be 1.47 KN.
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