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The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current

Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

$\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}$

$\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

If i = 2A and g = 10 cm

$\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97$

$f_{m}=-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N$

b) Using the coenergy of the system:

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2} }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N$

8 0

3 years ago

the highway department is to design a roadway over a swamp area. the swamp consists of very soft fine-grained soils obviously sa

garik1379 [7]

I'd like to propose that the embankment be built in stages. Because of the possibility of abrupt coloase of structure in swampy places. Soil carrying capacity is extremely low in those places. You must first densify the soil and make it acceptable using various procedures such as grouting, vibrification, restricting, and so on. These types of jobs are critical during the earliest stages of embankment development. After that, we can begin our construction work.

What is soil?

Soil, often known as earth or soil, is a complex mixture of organic matter, minerals, gases, liquids, and creatures that support life. Some scientific definitions distinguish dirt from soil by limiting the former term to displaced soil only. Soil is made up of two phases: a solid phase of minerals and organic matter and a porous phase that retains gases and water. As a result, soil is a three-state system composed of solids, liquids, and gases. Soil is the result of various elements interacting throughout time, including climate, relief, creatures, and the parent components of the soil.

To learn more about soil
brainly.com/question/28201561

#SPJ4

5 0

1 year ago

use engineering judgement to estimate the size of cooling and heating equipment that is needed for a three story, 31,000 square

Blizzard [7]

Answer: operating cost = 22820.736 $

Energy = 32.291 KBtu/sf.

Explanation:

Total heating load is given as = 31000 * 31.25

= 968.75* 103 Btu

From the cooling capacity application;

If 1000ftsquare = 2.8 TR

Therefore 30000ftsquare = x

Where x is the total cooling load,

Therefore x = (30000 * 2.8) / 1000

x = 84TR.

Therefore, the total cooling load, x = 84TR

Using conversion factor;

i.e. converting Btu/hr to MBH

1 MBH = 1000 Btu per hour

we have 968.75* 103 Btu/hr = 968.75 MBH

let us proceed.

Estimating the “Annual operating cost” we first calculate the maximum operating cost (Avg/Annum)

For cooling load:

Max. operatn. Cost = 14 * 84 = 1176$

For heating load:

The average heating loading hours = 1000 hrs.

Conversion to Btu/yr gives = 968.75* 103 * 1000

= 968.75* 106 Btu/yr.

Conversion of 968.75* 103Btu to Kwh gives

1Btu = 0.000293

968.75* 103 Btu = 283.84 Kwh

Therefore, average cost gives;

(8.04/100) $ = 1Kwh

X = 283.84*103

X = 22820.736 $

Operating cost = 22820.736 $

Energy use in KBtu/sf is given as

E = 968.75* 103 / 30000

E = 32.291 KBtu/sf.

8 0

4 years ago

Determine the atmospheric pressure at a location where the barometric reading is 760 mm Hg and the gravitational acceleration is

uysha [10]

Answer:

The atmospheric pressure is found to be $104.378kPa$

Explanation:

We know that pressure exerted by a standing column of fluid is calculated using the equation

$P=\rho _{fluid}\times gh$

In our case the pressure of the standing column of mercury is equal to the atmospheric pressure.

According to the given data we have

$\rho _{fluid}=14000kg/m^{3}$

$g=9.81m/s^{2}$

$h=760mm=0.76m$

Using the values in the equation above we calculate atmospheric pressure to be

$P_{atm}=14000\times 9.81\times 0.760\\\\=104.378kPa$

8 0

4 years ago

Consider a thin-walled cylindrical tube having a radius of 65 mm that is to be used to transport pressurized gas. (a) (10 points

AlladinOne [14]

Answer:

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

Explanation:

Minimum required thickness is the thickness of a material without corrosion allowance for each component based on the appropriate design that consider pressure, mechanical and structural loading.

Given that:

radius (r) = 65 mm = 65 × 10⁻³ m

Factor of safety (N) = 3.5

Inside presssure ( $P_{in}$ ) = 11 MPa