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Dmitriy789 [7]
3 years ago
12

A(n) _________ is a current greater than the equipment rated current or conductor ampacity, which is confined to the normal cond

uctive paths provided by the conductors and other equipment.
Engineering
1 answer:
exis [7]3 years ago
3 0

Answer:

overcurrent

Explanation:

It is any electrical current in excess of the nominal value indicated in the protection device, in the electrical equipment or in the current carrying capacity of a conductor. The overcurrent can be caused by an overload, a short circuit or a ground fault.

The overcurrent raises the operating temperature in the different elements of the electrical installation where this presentation is.

An overcurrent can be an overload or electric shock current.

<u>Overload</u>: the overload current is an excessive current in relation to the nominal operating current. It occurs in drivers and other components of a distribution system. Overloads are in most cases, more frequent between a range of one to six times the nominal current level. They are caused by temporary increases in current and occur when the motors start or when the transformers are energized.

Electric shock: as the name implies, a electric shock current is one that flows out of the normal conduction pathways. Electric shock or fault currents can be hundreds of times greater than the nominal operating current.

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Let m be an integer in the set {0,1,2,3,4,5,6,7,8, 9}, and consider the following problem: determine m by asking 3-way questions
kupik [55]

Answer:

Take any algorithm if that algorithm solves this problem it can be represented as a ternary decision tree. Therefore each question has at most three answers.

There are ten possible verdicts, the height of such kind of tree should satisfy

ℎ >= ⌈log3(10)⌉ = 3

Hence no such algorithm can ask less than three questions in the worst case.

---

b)

Each and every internal node represents a question asking whether m belongs to one of three possible subset of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} or not

For example 0123|456|789 represented the questionDoes “m: belongs to {0, 1, 2, 3}, to {4, 5, 6}, or to {7, 8, 9}?"

Verdicts are placed in brackets "[ ]"

Explanation:

decision tree is attached below

5 0
3 years ago
To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.8 times each persons weight - and t
Yuri [45]

Answer:

μ = 0.55

Explanation:

Given that

Normal weight = 1.8 x weight of person

N= 1.8 mg

We know that friction force Fr

Fr= μ N

μ=Coefficient of friction

N=Normal force

To find  μ We have to equate friction and gravity force

Fr= Wt

μ N = m g

μ  x 1.8 m g = m g

μ = 0.55

So the coefficient of friction will be 0.55.

5 0
3 years ago
The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm
aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

T_max =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=14.5MPa

T_{min} =\frac{T_c}{j}  = \frac{x}{y}  = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}

=10.3MPa

7 0
3 years ago
Entropy change is evaluated using Eq. 6.2a based on an internally reversible process. Can the entropy change between two states
Vadim26 [7]

Answer:

YES

Explanation:

Entropy is an extensive property of the system entropy change that value of entropy change can be determined for any process between the states whether reversible or not. i have attached the formula to calculate entropy change which is independent of whether the system is reversible or not and can be determined for any process.

4 0
4 years ago
A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries helium at 200 kPa
Aleks04 [339]

Answer: a 8143.71 kJ/kg

b 393.15 K

Explanation:

This system is an isobaric process in which there is no change in pressure a quasistatic process where a pressure distribution exists

a since no change in pressure =0 the system does work thus

FOR HELIUM  properties in standard thermodynamic chart

cv = 3.1 kJ/kgK

M = Molar mass = 4 kg/kmol

R = Universal gas constant = 8.314 kJ/kg K

cp ≈ cv +R /M = 3.1 + 8.314 /4 = 5.1785 kJ/kgK  

Cp = cp * M = 5.1785 kJ/kgK * 4 kg/kmol  = 20.714 kJ/kgkmol

T = 120  °C  to Kelvin = 120 + 273.15k = 393.15 K

W =n Cp ΔT = 1 kmol * 20.714 kJ/kg kmol* 393.15 K = 8143.71 kJ/kg

b convert T °C = T K thus 120 + 273.15 K = 393.15 K

P₁/T₁ = P₂/T₂

200 kPa/ 393.15 K = 200 kPa/T₂

T₂ = 200 kPa * 393.15 K/ 200 kPa = 393.15 K or 120 k

7 0
4 years ago
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