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2 years ago

State three characteristic of lines of magnetic flux

Engineering

2 answers:

stepladder [879]2 years ago

5 0

Answer:

1. The magnetic flux line form a closed loop.

2. The magnetic flux line repel each other.

3. The magnetic flux line never intersect.

xxMikexx [17]2 years ago

3 0

The field is tangent to the magnetic field line.
Field strength is proportional to the line density.
Field lines cannot cross.
Field lines are continuous loops.

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The device whose operation closely matches the way the clamp-on ammeter works is

Ivanshal [37]

Answer:

The answer is

C. Split phase motor

Explanation:

Clamp meters rely on the principle of magnetic induction to make non contact AC current measurements. Electric current flowing through a wire produces a magnetic field.

Which is similar to basic mode of operation of electric motor and split phase motor is a type of electric motor.

What is a a clamp on meter?

Clamp meters are electrical testers which have wide jaws that are able to clamp around an electrical conductor. Originally designed as a single purpose tool for measuring AC current, clamp meters now include inputs for accepting test leads and other probes that support a wide range of electrical measurements, the jaws of a clamp meter permit work in tight spaces and permits current measurements on live conductors without circuit interruption.

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3 years ago

Can i use two shunts and one meter

Lyrx [107]

Answer:

Explanation:

6 0

3 years ago

Read 2 more answers

A bus travels the 100 miles between A and B at 50 mi/h and then another 100 miles between B and C at 70 mi/h.

stira [4]

Answer:

c. less than 60 mi/h

Explanation:

To calculate the average speed of the bus, we need to calculate the total distance traveled by the bus, as well as the total time of travel of the bus.

Total Distance Traveled = S = 100 mi + 100 mi

S = 200 mi

Now, for total time, we calculate the times for both speeds from A to b and then B to C, separately and add them.

Total Time = t = Time from A to B + Time from B to C

t = (100 mi)/(50 mi/h) + (100 mi)(70 mi/h)

t = 2 h + 1.43 h

t = 3.43 h

Now, the average speed of bus will be given as:

Average Speed = V = S/t

V = 200 mi/3.43 h

V = 58.33 mi/h

It is clear from this answer that the correct option is:

c. less than 60 mi/h

7 0

3 years ago

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown

barxatty [35]

Answer:

The specific weight of unknown liquid is found to be 15 KN/m³

Explanation:

The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.

Total Pressure = Pressure of oil + Pressure of unknown liquid

65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)

65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)

(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²

(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m

Specific Weight of Unknown Liquid = 15 KN/m³

4 0

3 years ago

Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a

Ann [662]

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

8 0

3 years ago

State three characteristic of lines of magnetic flux​

State three characteristic of lines of magnetic flux