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2 years ago

¿Cuándo se formarán cristales en una mezcla que se está evaporando?

1 answer:

5 0

Answer - La cristalización ye un procesu químicu pol cual a partir d'un gas, un líquidu o una disolución, los iones, átomos o molécules establecen enllaces hasta formar una rede cristalina, la unidá básica d'un cristal. La cristalización emplegar con bastante frecuencia en química para purificar una sustancia sólida.

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How many different powerball combinations are there

svlad2 [7]

Answer:

There 11,238,513 different Powerball

4 0

2 years ago

g A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner

galben [10]

This question is not complete, the complete question is;

A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is $88 Mpam^\frac{1}{2}$ , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured

Answer:

length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

Explanation:

Given the data in the question;

vessel thickness = 6 cm

fracture toughness k = $88 Mpam^\frac{1}{2}$

yield strength = 1250 MPa

hoop stress equal = 300 MPa

we know that, the relation between fracture toughness and crack length is expressed as;

k = (1.1)(2/π)(r√(πa))

where k is the fracture toughness, r is hoop stress and a is length of crack

so we rearrange to find length of crack

a = 1/π[( k / 1.1(r)(2/π)]²

a = 1/π[( kπ / 1.1(r)(2)]²

so we substitute

a = 1/π [( 88π / 1.1(300)(2/π)]²

a = 1/π[ 0.1754596 ]

a = 0.05585 m

a = 0.05585 × 100 cm

a = 5.585 cm

so, length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

8 0

2 years ago

Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volum

ryzh [129]

Answer:

The Question is incomplete, the complete question is as follows:

Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volume of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F?

A six-lane rural freeway (three lanes in each direction) has regular weekday users and currently operates at maximum LOS C conditions. The base free-flow speed is 65 mi/h, lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and the interchange density is 0.25 per mile. The highway is one rolling terrain with 10% large trucks and buses (no recreational vehicles), and the peak-hour factor is 0.90. Determine the hourly volume for these conditions

Explanation:

Make the assumption Base continuous flow velocity (BFFS)= 65 mph. 

Pitch width= 11 ft.

Decrease in lane width pace,fLW= 1.9 mph.

Complete Lateral clearance= 4 ft. Lateral clearance speed reduction, fLC= 0.8 mph.

Complete Width of the Ramp= 0.25 mile.

Velocity reduction proportional to the ramp height, f ID= 0 mph.

Assume lane number to be = 3.

Reduction in speed corresponding to no. of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 65 – 1.9 – 0.8 – 3 – 0 = 59.3 mph

Peak Flow, V veh/hr

Peak-hour factor = 0.90

Trucks = 10%

Rolling Terrain

fHV = 1/ (1 + 0.10 (2.5-1)) = 1/1.15 = 0.8696

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = V/ (0.90*3*0.8696*1.0) = 0.426V veh/hr/ln

Average speed of vehicles, S = FFS = 59.3 mph

Level of service C

Density of LOS C lies between 18 - 25 veh/mi/ln

Maximum density = 25 veh/mi/ln

Density = Vp /S = 25

0.426V = 25 * 59.3

V = 3480 veh/hr

b) V = 5435 veh/hr

LOS dropping to F

Max density = 45 veh/mi/ln

Density = Vp/S = 45

Vp = 45 * 59.3 = 2668.5 veh/hr/ln

V/(PHF * n * fHV * fP) = 2668.5

fHV = 5435/(0.9*3*2668.5*1.0) = 0.754

1/(1+0.10 (ET -1)) = 0.754

ET = 4.26 ~ 3.5

For 4% upgrade and 10% trucks with ET = 3.5, length of the grade is Greater than 1.0 miles

6 0

3 years ago

Read 2 more answers

When it comes to making a good impression in a work setting, it does not apply to an initial contact, since both people are meet

Art [367]

Answer:

you have only seconds in which a person will accept or reject an employee or firm

Explanation:

First impression matters that's why when looking for employment with an organisation, lack of a tie for men may lead to automatic rejection. You have to be smart both intellectually and physically. Therefore, it means that you have only seconds in which a person will accept or reject an employee or firm.

4 0

3 years ago

A tank with a volume of 8 m3 containing 4 m3 of 20% (by volume) NaOH solution is to be purged by adding pure water at a rate of

lawyer [7]

Answer:

The time necessary to purge 95% of the NaOH is 0.38 h

Explanation:

Given:

vfpure water(i) = 3 m³/h

vNaOH = 4 m³

xNaOH = 0.2

vfpure water(f) = 2 m³/h

pwater = 1000 kg/m³

pNaOH = 1220 kg/m³

The mass flow rate of the water is = 3 * 1000 = 3000 kg/h

The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg

When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg

The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³

The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h

4 0

3 years ago