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3 years ago

Architects have identified the two most important knowledge areas for their profession as design and mathematics.

Engineering

1 answer:

a_sh-v [17]3 years ago

6 0

Answer:

False

Explanation:

The two most important fields of expertise for their career were not defined as architecture and mathematics by architects.

It offers expertise and expertise to architects and engineers that can not be gained by practicing alone.

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A water contains 50.40 mg/L as CaCO3 of carbon dioxide, 190.00 mg/L as CaCO3 of Ca2 and 55.00 mg/L as CaCO3 of Mg2 . All of the

Galina-37 [17]

Answer:

Total sludge = 123426kg/d

Explanation:

The reaction is given as;

H2Co3 + Ca(OH)2 ⇆ CaCo3 + 2H20

1 1 1 2 moles

Calculating the concentration of C02, we have

Concentration of C02 = concentration of CaCo3/Molecular weight of Caco3

= 50.4/100.09

= 0.5035mol/L

Sludge of Co2 = Conc. of Co2 * Q * MW of CaCo3 *10^-6

= 0.5035 * 253.6 *10^6 * 100.09 * 10^-6

= 12780kg/d

From the equation Ca2+ + 2HCo3- + Ca(OH)2 ⇄ 2CaCo3 + 2H2O

1 mole of calcium yields 2 moles of CaCo3

Therefore, Concentration of Ca2+ = Conc. of CaCo3/Mw of CaCO3

= 190-30/100.09

=1.599mol/L

Calculating sludge of calcium:

Sludge of Ca = 2 * Conc. of ca * Q * mw of CaCO3 * 10^-6

= 2 * 1.599 *253.6*10^6* 100.09 * 10^-6

= 811742kg/d

From the equation,

Mg2+ +2HCO3- + Ca(OH)2 ⇄ MgCO3 + 2CaCO3 + 2H2O

1 mole of mg yields 2 moles CaCO3 and 1 mole of Mg(OH)2

Concentration of Mg2+ = Conc, of CaCO3 /Mw of CaCo3

= 55- 10/100.09

= 0.4496mol/L

Sludge of Mg = 2 * Conc. of Mg * Q * mw of CaCO3 * 10^-6 +* Conc. of Mg * Q * mw of Mg(OH)2 * 10^-6

= 2 * 0.4496 * 253.5*10^6 * 100.09 * 10^-6 + 0.4996* 253.5*10^6 58.3 * 10^-6

= 29472kg/d

Total Sludge = Sludge of CO2 + Sludge of Ca + Sludge of Mg

12780+ 81174 + 29472

= 123426kg/d

6 0

3 years ago

Read 2 more answers

A binary liquid system exhibits LLE at 25°C. Determine from each of the following sets of miscibility data estimates for paramet

Phoenix [80]

Answer:

(a) - A12 = A21 = 2.747

(b) - A12 = 2.148; A21 = 2.781

(c)- A12 = 2.781; A21 = 2.148

Explanation:

(a) - x1(a) = 0.1 | x2(a) = 0.9 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

x1(a)*γ1(a) = x1(b)γ1(b)

x2(a)*γ2(a) = x2(b)γ2(b)

A12 = A21 = 2.747

(b) - x1(a) = 0.2 | x2(a) = 0.8 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

x1(a)*γ1(a) = x1(b)γ1(b)

x2(a)*γ2(a) = x2(b)γ2(b)

A12 = 2.148; A21 = 2.781

(c) - x1(a) = 0.1 | x2(a) = 0.9 | x1(b) = 0.8 | x2(b) = 0.2

LLE equations:

x1(a)*γ1(a) = x1(b)γ1(b)

x2(a)*γ2(a) = x2(b)γ2(b)

A12 = 2.781; A21 = 2.148

7 0

3 years ago

Block A has a weight of 8 lb. and block B has a weight of 6 lb. They rest on a surface for which the coefficient of kinetic fric

kkurt [141]

Answer:

For block A, a = 9.66 ft/s²

For block B, a = 15 ft/s²

Explanation:

A free body diagram for this force system is attached to this solution

Mass of block A = m₁ = 8 lb

Mass of block B = m₂ = 6 lb

Coefficient of kinetic friction = μ

Normal reaction on the blocks = N

Spring stiffness of the spring btw block A and B = k = 20 lb/ft

Compression of the spring = 0.2 ft

Analysing Block A first

The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 8 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 8/32.2 = 0.248 lbm

(20×0.2) - (0.2 × 8) = (0.248) aₓ

aₓ = 9.66 ft/s²

The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring

Sum of forces in the y-direction = 0

So, the weight of the block = Normal reaction of the surface on the block

N = W = 6 lb

Sum of forces in the x-direction = maₓ

(k × x) - (μ × N) = maₓ

m = W/g = 6/32.2 = 0.186 lbm

(20×0.2) - (0.2 × 6) = (0.186) aₓ

aₓ = 15 ft/s²

4 0

4 years ago

Chemical milling is used in an aircraft plant to create pockets in wing sections made of an aluminum alloy. The starting thickne

Lelu [443]

Answer:

a) metal removal rate is 1915.37 mm³/min

b) the time required to etch to the specified depth is 500 min or 8.333 hrs

Explanation:

Given the data in the question;

starting thickness of one work part of interest = 20 mm

depth of series of rectangular-shaped pockets = 12 mm

dimension of pocket = 200 mm by 400 mm

radius of corners of each rectangle = 15 mm

penetration rate = 0.024 mm/minute

etch factor = 1.75

To get the metal removal rate MRR;

The initial area will be smaller compare to the given dimensions of 200mm by 400mm and the metal removal rate would increase during the cut as area is increased. so'

A = 200 × 400 - ( 30 × 30 - ( π × 15² ) )

= 80000 - ( 900 - 707 )

= 80000 - 193

A = 79807 mm²

Hence, metal removal rate MRR = penetration rate × A

MRR = 0.024 mm/minute × 79807 mm²

MRR = 1915.37 mm³/min

Therefore, metal removal rate is 1915.37 mm³/min

b) To get the time required to etch to the specified depth;

Time to machine ( etch ) = depth of series of rectangular-shaped pockets / penetration rate

we substitute

Time to machine ( etch ) = 12 mm / 0.024 mm/minute

Time to machine ( etch ) = 500 min or 8.333 hrs

Therefore, the time required to etch to the specified depth is 500 min or 8.333 hrs

3 0

3 years ago

A fusible link should be how many wire sizes smaller than the actual circuit wire?

V125BC [204]

Answer:

.75mm

Explanation:

jak ammu chamba lng

5 0

3 years ago