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Margaret [11]
3 years ago
11

An inverting amplifier with +11 V supply voltages normally has a sinusoidal output of 10 Vpp. When checking the circuit with an

oscilloscope, you find that the output is 0 V. Which of the following could account for this problem? A) RF is shorted by a solder bridge.B) R1 is open. C) Vi = 0 D) All of the above

Engineering
1 answer:
Leto [7]3 years ago
5 0

Answer:

D) All of the above

Explanation:

To understand why we need to study case by case:

A) RF is shorted by a solder bridge: If RF is shorted, the output is going to be at the same potential that the inverting input due to a virtual short circuit (the inverting input is 0V because the non-inverting input is GND)

B) R1 is open: If R1 is open there is not input voltage to amplify therefore the output is 0V

C) Vi = 0: If Vi is equal zero there is not input voltage to amplify therefore the output is 0V

Note: Because there's is not schematic we assume the one in the picture down below, our logical explanation is complemented with a simulation that matches our results. For our case the amplifier requires a positive and negative supply, so we use an inverting amplifier with +11 V and -11V power supply.

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Bess [88]

Answer:

C/C++

Explanation:

C/C++

7 0
3 years ago
Calculate the viscosity(dynamic) and kinematic viscosity of airwhen
nikitadnepr [17]

Answer:

(a) dynamic viscosity = 1.812\times 10^{-5}Pa-sec

(b) kinematic viscosity = 1.4732\times 10^{-5}m^2/sec

Explanation:

We have given temperature T = 288.15 K

Density d=1.23kg/m^3

According to Sutherland's Formula  dynamic viscosity is given by

{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}, here

μ = dynamic viscosity in (Pa·s) at input temperature T,

\mu _0= reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

T_0 = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

\mu _0=4\pi \times 10^{-7}

T_0 = 291.15

\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-swhen T = 288.15 K

For kinematic viscosity :

\nu = \frac {\mu} {\rho}

kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec

3 0
3 years ago
A. A 3-kg plastic tank that has a volume of 0.2 m^3 is lled with liquid water. Assuming the density of water is 1000 kg=m^3, det
Tpy6a [65]

Answer:

The answer is below

Explanation:

a) The weight of the combined system is the sum of the weight of the water and the weight of the tank

m_{water}=V_{tank}.\rho_{wtaer}\\\\m_{water}=0.2m^3*1000kg/m^3\\\\m_{water}=200 \ kg\\\\m_{total} = m_{water}+m_{tank}\\\\But\ m_{tank}=3kg,therefore:\\\\m_{total} =200kg+3kg\\\\m_{total} =203\ kg\\\\weight_{total}=m_{total}g\\\\weight_{total}=203kg*9.81m/s^2\\\\weight_{total}=1991.43\ N

b) Since the weight of a system can be divided into smaller portions, hence weight is an extensive property.

c) When analyzing the acceleration of gases as they flow through a nozzle, the geometry of the nozzle which is an open system can be chosen as our system.

d) Given that:

\rho_{water}=1000kg/m^3\\\\1kg/m^3=0.062428lb/ft^3\\\\1000kg/m^3=1000kg/m^3*\frac{0.062428lb/ft^3}{kg/m^3}=62.43lb/ft^3\\ \\\rho=SG*\rho_{water}=1.03*62.43=64.272lb/ft^3\\\\P=P_{atm}+\rho g H\\\\P=14.7\ psia+64.272\ lb/ft^3*32.2\ ft/s^2*175\ ft*\frac{1\ ft^2}{12^2\ in^2}*\frac{1\ lbf}{32.2\ lbm.ft/s^2}  \\\\P=92.8\ psia

6 0
3 years ago
PLZZ HELP
9966 [12]

Answer:

Could ask a family member to help

Explanation:

5 0
3 years ago
Read 2 more answers
What does the air change rate represent?
Juli2301 [7.4K]

Answer and Explanation:

  • The removal or addition of air volume to the space is the air change rate
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  • The standard built home has a 0.5 to 1 of air change rate.
  • The rate of air change is dependent on the building (how the building form)  

3 0
3 years ago
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