A student wants to determine experimentally, without disconnecting any wires in the circuit, the DC current moving through a copper wire. Which of the following items of laboratory equipment would be sufficient to make the necessary measurements for this determination?(A) Magnetic field sensor only (B) Magnetic field sensor and meterstick (C) Bar magnet and meterstick (D) Stopwatch and meterstick (E) Voltmeter and current sensor

Answers

Answer 1

Answer:

A student wants to determine experimentally, without disconnecting any wires in the circuit, the DC current moving through a copper wire. Which of the following items of laboratory equipment would be sufficient to make the necessary measurements for this determination?

(E) Voltmeter and current sensor

Explanation:

DC (direct current) is the unidirectional stream or development of electric charge transporters (which are generally electrons). The power of the current can shift with time, yet the general direction of development remains the equivalent consistently. As a descriptor, the term DC is utilized in reference to voltage whose extremity never turns around. In a DC circuit, electrons rise up out of the negative, or less, shaft and move towards the positive, or additionally, post. All things considered, physicists characterize DC as making a trip from in addition to short.

A voltmeter is an instrument utilized for estimating electrical potential contrast between two focuses in an electric circuit. Simple voltmeters move a pointer over a scale in relation to the voltage of the circuit; advanced voltmeters give a numerical presentation of voltage by utilization of a simple to computerized converter. A voltmeter in a circuit outline is spoken to by the letter V around. Voltmeters are made in a wide scope of styles. Instruments for all time mounted in a board are utilized to screen generators or other fixed mechanical assembly. Versatile instruments, typically prepared to likewise gauge flow and obstruction as a multimeter, are standard test instruments utilized in electrical and hardware work. Any estimation that can be changed over to a voltage can be shown on a meter that is reasonably adjusted; for instance, pressure, temperature, stream or level in a compound procedure plant.

A current sensor is a gadget that identifies electric current in a wire and creates a sign relative to that current. The produced sign could be simple voltage or current or even a computerized yield. The produced sign can be then used to show the deliberate current in an ammeter, or can be put away for additional investigation in an information securing framework, or can be utilized with the end goal of control. Current sensors are either open-or shut circle. Open-circle current sensors measure air conditioning and DC currents and give electrical confinement between the circuit being estimated and the yield of the sensor (the essential current is estimated without electrical contact with the essential circuit, giving galvanic detachment). More affordable than their shut circle cousins, open-circle current sensors are commonly favored in battery-controlled circuits given their low-working force prerequisites and little impression highlights.


Related Questions

A cylindrical rod of 1040 Steel originally 10.5 mm in diameter is to be cold worked by drawing to a final diameter of 8.5 mm. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 750 MPa and a ductility of at least 12 %EL are desired. Explain, with details and calculations, how this may be accomplished.

Answers

Answer:

Check the explanation

Explanation:

The desired Cold working (which is any process or procedure of metalworking in which the metal body is formed at a lower temperature beyond its recrystallization, typically at the ambient temperature.) tensile strength in excess of 750 MPa and a ductility of at least 12 %EL can be seen in the attached image below.

Solar energy stored in large bodies of water, called solar ponds, is being used to generate electricity. If such a solar power plant has an efficiency of 3 percent and a net power output of 150 kW, determine the average value of the required solar energy collection rate, in Btu/h.

Answers

Answer:

17060700 btu/h

Explanation:

Power output from power plant = 150 kW.

This power output is at 3% efficiency, this means power available in pond is 0.30P

Now, 150 kW = 0.03P

P = 150/0.03 = 5000 kW of power in the pond.

From basic conversion,

1 kW = 3412.14 btu/h

5000 kW = 5000 x 3412.14

average value of the required solar energy collection rate equal

17060700 btu/h

Solar pond power plant with 3% efficiency and 150 kW output needs ~17,060,000 Btu/h solar energy collection rate.

To find the average value of the required solar energy collection rate in Btu/h, we need to calculate the total solar energy collected per unit time and then convert it into Btu/h.

First, let's find the total solar energy collected per unit time:

Given:

- Net power output = 150 kW

- Efficiency = 3%

We know that efficiency is the ratio of output power to input power, so:

Efficiency = (Net Power Output / Solar Energy Input) * 100

Rearranging the equation to find the solar energy input:

Solar Energy Input = (Net Power Output / Efficiency) * 100

Now, let's substitute the given values:

Solar Energy Input = (150 kW / 3%) * 100

Solar Energy Input = (150,000 W / 0.03) * 100

Solar Energy Input = 5,000,000 W

Now, we convert this into Btu/h. To do this, we'll use the conversion factor:

1 watt = 3.412 Btu/h

So, the solar energy input in Btu/h will be:

Solar Energy Input (Btu/h) = (Solar Energy Input in watts) * (3.412 Btu/h per watt)

Solar Energy Input (Btu/h) = 5,000,000 W * 3.412 Btu/h per watt

Solar Energy Input (Btu/h) ≈ 17,060,000 Btu/h

Therefore, the average value of the required solar energy collection rate is approximately 17,060,000 Btu/h.

public class Point { private int x; private int y; public void setx(int x) this.x = x; public void setY(int y) this.y=y; public void setXY(int x, int y) X=y; ysy; public boolean IsEqual(Point p) return ((p.x == x) && (p.y == y)); public void DisplayPoint() System.out.println("(" + x + " +X+ + y + ")"); public void movePoint(int deltax, int deltay) {!! 'x' is increased by deltax and y is increased by deltay public void SetLocation(Point P) {//make point to have the specified location P. public static void main(String[] args) { // TODO Auto-generated method stub Point P1 = new Point(); Point P2 = new Point(); P1.setx(5); P1.setY(6); P2.setx(5); P2.setY(6); System.out.println(P1.IsEqual(P2)); P2.setXY(3,4); System.out.println(P1.IsEqual(P2)); Given the class 'Point' which has two member variables x, and y. What are x and y for Point object P1 after the end of the main method ? (Pay close attention to the setXY method) x=6, y=5 Ox=6, y= 6 x=5, y= 6 x=5, y= 5

Answers

Answer:

x =5

y =6

Explanation:

The reason behind that is that while using the reference P2 we have setted the values of X and Y both as 5 and 6 respectively in the main function.

After setting we are calling the setXY function which sets the value to 3 and 4 for X and Y respectively but does not implement in the value to the instance.

Apart from which setX and setY function are setting the values with respect to the instance variable using "this" keyword.

Hence, the values 5 and 6 are been displayed at the end of the main method.

A museum has three rooms, each with a motion sensor (m0, m1, and m2) that outputs 1 when motion is detected. At night, the only person in the museum is one security guard who walks from room to room. Using the combinational design process, create a circuit that sounds an alarm (by setting an output A to 1) if motion is ever detected in more than one room at a time (i.e., in two or three rooms), meaning there must be an intruder or intruders in the museum.

a. Describe the behavior of this problem with a truth table.

b. Find the minimal logic expression based on the truth table. You can use either K-map or Boolean algebra.

c. Construct the gate-level circuit schematic using the logic expression derived in problem b.

Answers

Answer:

a) see attachment

b) A= m0m1+ m1m2+ m0m2

see attachment for K-map

c) see attachment

Explanation:

a) see attachment for truth table

b) see attachment for k-map

A= m0m1+ m1m2+ m0m2

c) see attachment for gate level circuit

A large plate is at rest in water at 15?C. The plate is suddenly translated parallel to itself, at 1.5 m/s. The resulting fluid movement is not exactly like that in a b.l. because the velocity profile builds up uniformly, all over, instead of from an edge. The governing transient momentum equation, Du/Dt = ?(?2u/?y2), takes the form 1 ? ?u ?t = ?2u ?y2


`Determine u at 0.015 m from the plate for t = 1, 10, and 1000 s. Do this by first posing the problem fully and then comparing it with the solution in Section 5.6. [u 0.003 m/s after 10 s.]

Answers

Answer:

Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s

   

Explanation:

du/u² = dt = ∫du/2.3183 = ∫dt

0.4313 u = t + c

(a) t = 0, u= 15m/s, c = 0.647

u = t+c/0.4313 = t + 0.647/0.4313

(a) when t= 1   u = 1+ 0.647/0.4313 = 3.8187m/s

(b) when t= 10   u = 10 + 0.647/0.4313 = 24.0858m/s

(c)when t= 1000  u = 1000 + 0.647/0.4313 = 3220.071m/s

Air enters the compressor of a cold air-standard Brayton cycle with regeneration at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rate of entropy production in the regenerator, in kW/K

Answers

(a) The thermal efficiency of the cycle.

The thermal efficiency of the cycle is given by
NThermal= net work/ heat supplied
NThermal= WNet/QH
=2026.08/4547.82= 44.55%

(b) The backwork ratio:
compressor work/ turbine work
=1760.76/3786.84= 0.465

(c) The net power developed:
WNet= WT-WC
WNet= 3786.84- 1760.76= 2026.08 kW

For some material, the heat capacity at constant volume Cv at 34 K is 0.81 J/mol-K, and the Debye temperature is 306 K. Estimate the heat capacity (in J/mol-K) (a) at 44 K, and (b) at 477 K.

Answers

Answer:

a. Heat Capacity = 1.756J/mol-K

b. Heat Capacity = 24.942J/mol-k

Explanation:

Given

Constant volume Cv = 0.81J/mol-k

T1 = 34K

Td = Debye temperature = 306 K. Estimate the heat capacity (in J/mol-K) a. 44 K

First, The value of the temperature-independent constant.

Using Cv = AT³

Make A the subject of formula

A = Cv/T³

Substitute each values

A = 0.81/34³

A = 0.000020608589456543

A = 2.061 * 10^-5J/mol-k

The heat capacity changes with the temperature; below is the relationship between heat capacity and the temperature

Cv = AT³

So, The heat capacity when T = 44k is then calculated as

Cv = 2.061 * 10^-5 * 44³

Cv = 1.755522084266232

Cv = 1.756J/mol-K

(b) at 477 K.

Because the temperature is larger than the Debye temperature, the specific heat is calculated using as:

Cv = 3R

Where R = universal gas constant

R = 8.314J/mol-k

Cv = 3 * 8.314

Cv = 24.942J/mol-k

Answer:

a) 1.75 J/mol-k

b) 24.94 J/mol-k

Explanation:

We are given:

Cv = 0.81J/mol-k

Debye temperature = 306k

Heat capacity varies with temperature and the relationship between them is given as:

Cv = AT³

To find temperature independent constant A, we have:

[tex] A = \frac {C_v}{T^3} [/tex]

[tex] A = \frac{0.81J/mol-k}{34^3} [/tex]

[tex] A = 2.06*10^-^5 [/tex]

a) for temperature at 44k

Cv = AT³

[tex] C_v = 2.06*10^-^5 * 44^3 [/tex]

Cv = 1.75 J/mol-k

b) for T at 477k

Since the temperature 447k is higher than Debye's temperature, the specific heat will be given as:

Cv = 3R

Where R = universal gas constant = 8.314 J/mol-k

Therefore,

Cv = 3 * 8.314 J/mol-k

Cv = 24.94 J/mol-k

This code initially states that n=10 and f=n. Once the code enters the while loop, it stays in the while loop until the condition n>1 is not true. n is subtracted by 1 every iteration. This means that f = 10*9*8*7*6*4*3*2*1. With 9,8,7,6,5,4,3,2, and 1 being repeated n values. The disp (f) command displays the value of f after the while loop, which is equal to (n!).

Answers

Answer:

n = 10; f = n; while(n > 1)    n = n - 1;    f = f * n; end disp(f);

Explanation:

The solution code is written in Matlab.

Firstly, we initialize n to 10 and set n to f variable (Line 1-2).

Next, create a while loop that will continue to loop until n is equal or smaller than one. In the loop, decrement n by one and multiply n with current f_variable.

At last display value of f. The output is  3628800.

Carbon dioxide used as a natural refrigerant flows through a cooler at 10 MPa, which is supercritical, so no condensation occurs. The inlet is at 220°C and the exit is at 50°C. Find the specific heat transfer.

Answers

Answer:

The answer which is a calculation can be found as an attached document

Explanation:

Answer:

Answer is -286.78 kJ/kg

Refer below.

Explanation:

Refer to the pictures for brief explanation.

A pipe of 10 cm inner diameter is used to send crude oil over distance of 400 meters. The entire pipe was laid horizontal. The viscosity of the oil is 10 cp. The density of oil is 800 kg/m3 . a. (20 pts) The desired volumetric flow rate is 0.1 m3 /min. What is the Reynolds number of this flow? Is the flow laminar or turbulent? What is the pressure difference needed to generate this flow rate? b. (15 pts) Three month later, the operator found that they had to triple the pressure difference to maintain the flow rate at 0.1 m3 /min. The operator thought that wax deposition from the oil had reduced the inner diameter of the tube. Based on this assumption, can you estimate the reduced inner diameter of the tube? What is the Reynolds number of the flow?

Answers

Answer:

See explaination

Explanation:

Looking at Reynolds number, we can go ahead and describe the Reynolds number as a dimensionless value that is used to determine whether the fluid is exhibiting laminar flow (R less than 2300) or turbulent flow (R greater than 4000). Laminar flow is when a fluid moves smoothly and is predictable.

Check attachments

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