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kvasek [131]
3 years ago
15

Create a simulation on multisim of over-voltage & under-voltage protection circuit.

Engineering
1 answer:
Eduardwww [97]3 years ago
6 0

Answer:protection circuit overload as usual

circuit overload

Explanation:

it is ery easy for wverone

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A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he
hichkok12 [17]

Answer:

ΔT=  11.94 °C

Explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400  KJ

Heat input to the water = 1  KW

Heat input the water= 1 x 15 x 60

                                =900 KJ

By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

C_p=4.187 \frac{KJ}{kg-K}

Q=mC_p\Delta T

Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So  rise in temperature of water ΔT=  11.94 °C

6 0
3 years ago
A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3
labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass co_2 = 1 kg

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of co_2 = 44

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

V_1 = \frac{m RT_1}{P_1}

       =\frac{1*0.189*298}{104}

V_1 = 0.5415 m3

V_2 = \frac{m RT_2}{P_2}

     =\frac{1*0.189*584}{1068}

V_1 = 0.1033 m3

WORK DONE

W =P_{avg}*{V_2-V_1}

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

\Delta U  = m *c_v*{V_2-V_1}

\Delta U  = 1*0.657*(584-298)

\Delta U  =187.902 kJ

HEAT TRANSFER

Q = \Delta U  +W

   = 187.902 +(-256.46)

Q = -68.859 kJ

7 0
3 years ago
When you hover over an edge or point, you are activating ____________ in SketchUp?
7nadin3 [17]

Answer:b

hope thiss helps

Explanation

I took the quiz

7 0
2 years ago
Read 2 more answers
A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?​
dexar [7]

Answer:

8.85 Ω

Explanation:

Resistance of a wire is:

R = ρL/A

where ρ is resistivity of the material,

L is the length of the wire,

and A is the cross sectional area.

For a round wire, A = πr² = ¼πd².

For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.

Given L = 500 ft and d = 0.03 in = 0.0025 ft:

R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)

R = 8.85 Ω

5 0
3 years ago
The high electrical conductivity of copper is an important design factor that helps improve the energy efficiency of electric mo
ludmilkaskok [199]

Answer:

B

Explanation:

This is a two sample t-test and not a matched pair t-test

null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors

alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.

So, option D is rejected

The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.

Option A and C are also rejected

5 0
3 years ago
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