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2 years ago

Voltage drop testing is being discussed.

Engineering

1 answer:

Airida [17]2 years ago

8 0

Answer:

Technician B only is correct

Explanation:

Voltage drop testing is a method used to find the amount of electrical resistance available in an high amperage circuit that involves connecting the leads of the meter in parallel to the circuits being tested such that disassembly is not required

In voltage drop test, the red voltmeter lead and black voltmeter lead are placed at two points on the same side of the circuit connection such that the leads are in between two positive connection or two negative connection (rather than connecting the red to the positive and the black to the negative sides of the circuit) and digital voltmeter is used for the voltage drop measurement across the lead while the connection is under load.

Therefore, Technician B only is correct

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Each cout statement has a syntax error. Type the first cout statement, and press Run to observe the error message. Fix the error

saul85 [17]

Answer:

1. cout << "Num: " << songNum << endl;

2. cout << songNum << endl;

3. cout << songNum <<" songs" << endl;

Explanation:

//Full Code

#include <iostream>

using namespace std;

int main ()

{

int songNum;

songNum = 5;

cout << "Num: " << songNum << endl;

cout << songNum << endl;

cout << songNum <<" songs" << endl;

return 0;

}

1. The error in the first cout statement is that variable songnum is not declared.

C++ is a case sensitive programme language; it treats upper case and lower case characters differently.

Variable songNum was declared; not songnum.

2. Cout us used to print a Variable that has already been declared.

The error arises in int songNum in the second cout statement.

3. When printing more than one variables or values, they must be separated with <<

4 0

3 years ago

The purpose of pasteurizing milk is to A. Kill pathogens B. Break down milk fat C. Add vitamins and minerals D. Prevent spoilage

motikmotik

Answer: A Kill pathogens

7 0

2 years ago

Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the

alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

$\dot Q_{ref} - \dot Q_{w} = 0$

$\dot Q_{ref} = \dot Q_{w}$

$\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})$

The mass flow rate of the cooling water is now cleared:

$\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}$

Given that $h_{ref,in} = 808.34\,\frac{kJ}{kg}$ , $h_{ref, out} = 88.82\,\frac{kJ}{kg}$ , $h_{w,out} = 104.83\,\frac{kJ}{kg}$ and $h_{w,in} = 62.98\,\frac{kJ}{kg}$ , the mass flow of the cooling water is:

$\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)$

$\dot m_{w} = 123.788\,\frac{kg}{min}$

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

4 0

3 years ago

Describe the energy transformation that occurs when a television is used

Morgarella [4.7K]

A television transforms electric energy into sound energy and radiant energy.

<h3>What is energy transformation?</h3>

Energy transformations are processes that convert energy from one type into another. Any type of energy use must involve some sort of energy transformation.

Some examples of energy transformations are, A television converts electric energy into sound and light energy. Lightbulbs convert electric energy into light energy. A hair dryer converts electrical energy into thermal and sound energy.

Learn more about Energy transformations here,

brainly.com/question/8822715

#SPJ4

6 0

2 years ago

Which of the following are TRUE concerning rectifier circuits? Select all that apply.

Dovator [93]

Explanation:

a. The output of an ideal full wave rectifier is zero volts only when the input is zero volts.

True

The output of an ideal full wave rectifier is zero volts only when the input is zero since 1st diode is forward biased during one half cycle of the input and 2nd diode is forward biased during the other half cycle of the input therefore, it fully utilizes both the input cycles so the output voltage is only zero when the input is zero.

b. Half-wave rectifier circuits need a minimum of 4 diodes to operate.

False

A Half-wave rectifier circuits need a minimum of 1 diode to operate, whereas a full-wave bridge rectifier need minimum of 4 diodes to operate.

c. In an ideal full wave bridge rectifier, half of the diodes are in the ON state and half of the diodes are in the OFF state at any given time the input voltage is not zero.

True

A full-wave bridge rectifier consists of 4 diodes, where 2 diodes are functional in half of the cycle(so the other 2 are off) and other 2 diodes are functional in the other half cycle( so the other 2 are off).

d. The output of an ideal half wave rectifier is zero volts only when the input is zero volts.

False

The output of an ideal half wave rectifier is zero during half of the cycle when the diode is reversed biased and doesn't conduct even though input voltage is not zero volts at this point.

e. A turn-on voltage of a diode (y,) greater than zero can cause the output of a full wave rectifier to be zero volts even when the input is not zero volts.

False

A turn-on voltage of a diode (y,) greater than zero cannot cause the output of a full wave rectifier to be zero rather there will be a little voltage drop across the output of full wave rectifier due to this turn-on voltage of diode which is usually 0.7 volts for silicon based diodes.

f. An advantage of the half wave rectifier is that is can use a smoothing capacitor, while a full wave rectifier cannot.

False

Smoothing capacitor can be used in both half wave rectifier as well as full wave rectifier.

8 0

3 years ago