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3 years ago

_____ are used to control the flow of electricity in a circuit.

Engineering

2 answers:

Travka [436]3 years ago

8 0

Answer:

Switches control the flow of electricity in a circuit.

Cloud [144]3 years ago

5 0

Answer:

switches

Explanation:

switches because it a device used to control electric circuit regarding whether current should flow or not.

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The status of which of these determines the sequence in which output devices, such as solenoid values and motor contactors, are

Mkey [24]

A. Physical I/O sensors

Safety switches, operator inputs, travel limit switches etc

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3 years ago

Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arra

VashaNatasha [74]

Answer:

Explanation:

ADT for an 2-D array:

struct array{

int arr[10];

}arrmain[10];

An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :

1) struct array{

int *p;

}arr[1000];

2) struct array{

int *p;

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3 years ago

Sinks must be used for the correct intended purpose to prevent

irakobra [83]

Answer:

... spilling water or getting anything cascading onto the floor

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3 years ago

214Bi83 --> 214Po84 + eBismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle,

Zolol [24]

Answer:

Explanation:

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3 0

3 years ago

Read 2 more answers

An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter

slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg , 1980.39 MJ/Kmol

Explanation:

molar mass of ethanol (C2H5OH) = 46 g/mol

molar mass of octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol = 0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV of ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol) for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV in MJ/kg = 0.85 * 29.7 + 0.15 * 47.9 = 32.43 MJ/kg

HHV in MJ /kmol = 0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0

2 years ago