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Umnica [9.8K]
2 years ago
5

Does anyone know the answer to this

Engineering
2 answers:
soldi70 [24.7K]2 years ago
7 0

Answer:

A

Explanation:

Liula [17]2 years ago
6 0

Answer:

A

Explanation:

The fuse breaks the circuit if a fault in an appliance causes too much current to flow. The fuse contains a piece of wire that melts easily. If the current going through the fuse is too great,the wire heats up until it melts and breaks the circuit

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A heat engine that receives heat from a furnace at 1200°C and rejects waste heat to a river at 20°C has a thermal efficiency of
viktelen [127]

Answer:

second-law efficiency  = 62.42 %

Explanation:

given data

temperature T1 = 1200°C = 1473 K

temperature T2 = 20°C  =  293 K

thermal efficiency η = 50 percent

solution

as we know that thermal efficiency of reversible heat engine between same  temp reservoir

so here

efficiency ( reversible ) η1 = 1 - \frac{T2}{T1}      ............1

efficiency ( reversible ) η1  = 1 - \frac{293}{1473}  

so efficiency ( reversible ) η1  = 0.801

so here second-law efficiency of this power plant is

second-law efficiency = \frac{thernal\ efficiency}{0.801}

second-law efficiency = \frac{50}{0.801}  

second-law efficiency  = 62.42 %

3 0
3 years ago
In the circuit given below, R1 = 17 kΩ, R2 = 74 kΩ, and R3 = 5 MΩ. Calculate the gain 1formula58.mml when the switch is in posit
Elenna [48]

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)

3 0
3 years ago
Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo
rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

KE = 300 KJ

Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

7 0
3 years ago
The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc
Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy  

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.  

Given Data  

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist  

∅=4°

∅=4*\pi/180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist  

∅=TL/GJ

∅=TL/G*\pi/2*c^4

∅=2TL/G*\pi*c^4

c=\sqrt[4]{2TL/\pi G }∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J*\pi/2*c^4)]*c

г =[2T/(J*\pi*c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required  

We will use larger of the two values  

c= 18.0569 x 10^-3 m  

c = 18.0569 mm  

3 0
3 years ago
What type of test can show a chemical engineer if a material remains in a system and accumulates or if it moves right through?
UkoKoshka [18]

A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.

<h3>What is a chemical engineer?</h3>
  • Processes for manufacturing chemicals are created and designed by chemical engineers.
  • To solve issues involving the manufacture or usage of chemicals, fuel, medications, food, and many other goods, chemical engineers use the concepts of chemistry, biology, physics, and math.
  • A wide range of sectors, including petrochemicals and energy in general, polymers, sophisticated materials, microelectronics, pharmaceuticals, biotechnology, foods, paper, dyes, and fertilizers, have a significant demand for chemical engineers.
  • Chemical engineering is undoubtedly difficult because it requires a lot of physics and math, as well as a significant number of exams at the degree level.

To learn more about chemical engineer, refer to:

brainly.com/question/23542721

#SPJ4

7 0
2 years ago
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