If the symbol is on the top and bottom of the reference line, we call this _________

Do the coil resistances have any effect on the plots?

PolarNik [594]

Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.

8 0

3 years ago

Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r

My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0

4 years ago

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

3 years ago

Which of the following hazards is shared by surface mining and sub-surface mining?

lubasha [3.4K]

B I guess. since they both have potential to collapse

4 0

3 years ago

Read 2 more answers

A manometer is used to measure the air pressure in a tanlc The fluid used has a specific gravity of 1.25, and the differentialhe

Alenkasestr [34]

Answer:

(a) 11.437 psia

(b) 13.963 psia

Explanation:

The pressure exerted by a fluid can be estimated by multiplying the density of the fluid, acceleration due to gravity and the depth of the fluid. To determine the fluid density, we have:

fluid density = specific gravity * density of water = 1.25 * 62.4 lbm/ft^3 = 78 lbm/ft^3

height = 28 in * (1 ft/12 in) = 2.33 ft

acceleration due to gravity = 32.174 ft/s^2

The change in pressure = fluid density*acceleration due to gravity*height = 78*32.174*(28/12) = 5855.668 lbm*ft/(s^2 * ft^2) = 5855.668 lbf/ft^2

The we convert from lbf/ft^2 to psi:

(5855.668/32.174)*0.00694 psi = 1.263 psi

(a) pressure = atmospheric pressure - change in pressure = 12.7 - 1.263 = 11.437 psia

(b) pressure = atmospheric pressure + change in pressure = 12.7 + 1.263 = 13.963 psia

8 0

3 years ago

If the symbol is on the top and bottom of the reference line, we call this __________ sides.