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liberstina [14]
3 years ago
11

The following two questions refer to the circuit below. Consider a non-ideal op amp where the output can saturate. The open loop

gain A = 2 x 10^{4}10 4 , where v_{o}v o ​ =−Av_{s}v s ​ . The positive supply voltage for the op-amp is +V_S = 15+V S ​ =15V. The negative supply voltage for the op-amp is -V_S = -10−V S ​ =−10V. What is the most positive value v_{s}v s ​ can take before the amplifier saturates? Express your answer in mV and omit units from your answer.

Engineering
1 answer:
kotykmax [81]3 years ago
6 0

Answer:

The most positive is value of Vs is 0.5mV

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The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,
babunello [35]

Answer:

Qx = 9.109.10^5 \times 10^{-6} m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × 10^6 Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 \times 10^{-3} )²× 9  \times 10^{-3}  × sin18 × cos18

Qd = 94.305 × 10^{-6} m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × 10^{6} × π × 85 \times 10^{-3}  × ( 9  \times 10^{-3} )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × 10^{-6} m³/s

so here

volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × 10^{-6}  - 85.2 × 10^{-6}  

Qx = 9.109.10^5 \times 10^{-6} m³/s  

8 0
3 years ago
Steam at 1 MPa, 300 C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. The mass flow rate of this steam i
stealth61 [152]

Answer:

\dot m = 2.74 kg/s

Explanation:

given data:

pressure 1 MPa

diameter of pipe  =  30 cm

average velocity = 10 m/s

area of pipe= \frac[\pi}{4}d^2

                 = \frac{\pi}{4} 0.3^2

A = 0.070 m2

WE KNOW THAT mass flow rate is given as

\dot m = \rho A v

for pressure 1 MPa, the density of steam is = 4.068 kg/m3

therefore we have

\dot m = 4.068 * 0.070* 10

\dot m = 2.74 kg/s

7 0
4 years ago
The welding method that requires the operator to observe and only make corrections is
SVEN [57.7K]
Automatic manual semiautomatic
3 0
3 years ago
Read 2 more answers
Before cutting coarse screw threads, the operator should lubricate: A. The leadscrew and gearbox B. The ways and cross slide C.
Maksim231197 [3]

Answer:

(d) all of the above

Explanation:

before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial

so option (D) will be correct because we need lubricate in all the given parts  

8 0
3 years ago
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas
jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine (\eta), no unit, is defined by this formula:

\eta = \frac{\dot W}{\dot Q} (1)

Where:

\dot Q - Heat input, in watts.

\dot W - Power output, in watts.

If we know that \dot W = 600\,W and \eta = 0.3, then the heat input from the combustion phase is:

\eta = \frac{\dot W}{\dot Q}

\dot Q = \frac{\dot W}{\eta}

\dot Q = \frac{600\,W}{0.3}

\dot Q = 2000\,W

The heat input from the combustion phase is 2000 watts.

8 0
3 years ago
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