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3 years ago

8

An amplifier which needs a high input resistance and a high output resistance is : Select one: a. A voltage amplifier b. None of

these c. A transresistance amplifier d. A current amplifier e. A transconductance amplifier Clear my choice

1 answer:

True [87]3 years ago

3 0

Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

Voltage amplifier needs high input and low output resistance.

Current amplifier needs Low Input and High Output resistance.

Trans-conductance amplifier Low Input and High Output resistance.

Trans-Resistance amplifier requires High Input and Low output resistance.

Therefore, the correct answer is "None of these "

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A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler an

konstantin123 [22]

Answer:

The right solution is "28.45%".

Explanation:

The given values are:

$P_4=50\ kPa$

$h_4=0.7(2304.7)+340.5$

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and,

$P_3=15 \ mPa$

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$s_3=sg$

$=5.3108 \ KJ/Kgh$

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$=0.66$

At $P_4=50 \ Kpa$ ,

$h_f=340.54$

or,

$V_f=0.001030 \ m^3/Kg$

then,

⇒ $h_2=340.54+0.001030(15\times 10^{3}-50)$

$=355.94 \ kJ/kg$

hence,

The isentropic efficiency of turbine will be:

⇒ $n_T=\frac{h_3-h_4}{h_3-h_{45}}$

$=\frac{2610.8-1953.83}{2610.8-1836.26}$

$=84.818$ (%)

The thermal efficiency of cycle will be:

⇒ $n_C=\frac{W_T-W_P}{2_{in}}$

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3 years ago

Identify five safety hazards that should be included in the design of the school

erik [133]

Answer:

1) function of fire doors and making sure theyre properly wired to fire alarms

2) proper water piping and purifaction for water fountains and sinks.

3) falty sprinkler systems/rsuty sprikler systems that wont work

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Explanation:

HOPE THIS HELPS good luck!!

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3 years ago

For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in.

makkiz [27]

Answer:

nP ≈ 4.9

nL = 1.50

Explanation:

GIVEN DATA

external load applied (p) = 85 kips

bolt stiffness ( Kb ) = 3(10^6) Ibf / in

Member stiffness (Km) = 12(10^6) Ibf / in

Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8

Number of bolts = 6

assumptions

for unified screw threads UNC and UNF

tensile stress area ( A ) = 0.1419 in^2

SAE specifications for steel bolts for grade 8

we have

Minimum proff strength ( Sp) = 120 kpsi

Minimum tensile strength (St) = 150 Kpsi

Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips

Given the following values

Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip

Preload stress

αi = 0.75Sp = 0.75 * 120 = 90 kpsi

stiffness constant

C = $\frac{Kb}{Kb + Km}$ = $\frac{3}{3+2}$ = 0.2

A) yielding factor of safety

nP = $\frac{sPAt}{Cp + Fi}$ = $\frac{120* 0.1419}{0.2*14.17 + 12.771}$

nP = 77.028 / 15.605 = 4.94 ≈ 4.9

B) Determine the overload factor safety

$nL = \frac{SpAt - Fi}{CP}$ = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17

= 17.028 - 12.771 / 2.834

= 1.50

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