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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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6.28 A six-lane freeway (three lanes in each direction) in rolling terrain has 10-ft lanes and obstructions 4 ft from the right

dimulka [17.4K]

Answer:

Assume Base free flow speed (BFFS) = 70 mph

Lane width = 10 ft

Reduction in speed corresponding to lane width, fLW = 6.6 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, fLC = 0.8 mph

Interchanges/Ramps = 9/ 6 miles = 1.5 /mile

Reduction in speed corresponding to Interchanges/ramps, fID = 5 mph

No. of lanes = 3

Reduction in speed corresponding to number of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 70 – 6.6 – 0.8 – 3 – 5 = 54.6 mph

Peak Flow, V = 2000 veh/hr

Peak 15-min flow = 600 veh

Peak-hour factor = 2000/ (4*600) = 0.83

Trucks and Buses = 12 %

RVs = 6 %

Rolling Terrain

fHV = 1/ (1 + 0.12 (2.5-1) + 0.06 (2.0-1)) = 1/1.24 = 0.806

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 2000/ (0.83*3*0.806*1.0) = 996.54 ~ 997 veh/hr/ln

Vp < (3400 – 30 FFS)

S = FFS

S = 54.6 mph

Density = Vp/S = (997) / (54.6) = 18.26 veh/mi/ln

7 0

3 years ago

In a 5V system if you were asked to take one input HIGH and another LOW what would you do (i.e. where would you connect them)?

dangina [55]

Answer:

HIGH from the supply voltage

LOW from ground

Explanation:

The answer depends on the kind of system and the purpose of the signal. But for practical reasons, in a DIGITAL system where 5V is HIGH and 0 V is LOW, 5 volts can be taken from the supply voltage (usually the same as high, BUT must be verified), and the LOW signal from ground.

If the user has a multimeter, it must be set to continuous voltage on 0 to 20 V range. Then place the probe in the ground of the circuit (must be a big copper area). Finally leave one probe in the circuit ground and place the other probe in some test points to identify 5 v.

4 0

3 years ago

A very long pipe of 0.05 m (r1) radius and 0.03 m thickness (r2 - r1) is buried at a depth of 2m (z) to transport liquid nitroge

Elenna [48]

Answer:

The surface temperature of the ground is = 296.946K

Explanation:

Solution

Given

r₁= 0.05m

r₂= 0.08m

Tn =Ti = 77K

Ki = 0.0035 Wm-1K-1

Kg = 1 Wm-1K-1

Z= 2m

Now,

The outer type temperature (Skin temperature pipe)

Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi T0 -T1/ln (r2/r1)

Thus,

10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05

⇒ T₀ -77 = 231.72

T₀= 290.72K

The shape factor between the cylinder and he ground

S = 2πL/ln 4z/D

where L = length of pipe

D = outer layer of pipe

S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m

The heat gained in the pipe is = S * Kg * (Tg- T₀)

(10* 1) = 1.606 * 1* (Tg- 290.72)

Tg - 290.72 = 6.2266

Tg = 296.946K

Therefore the surface temperature to the ground is 296.946K

6 0

2 years ago

Air is contained inside a vertical piston-cylinder assembly by a piston of mass 105 kg and face area of 0.05 m2 . The mass of ai

alexgriva [62]

Answer:

ΔQ = 1.06 KJ

Explanation:

The amount of heat transfer between the piston-cylinder system and the surrounding can easily be found by using the First Law of Thermodynamics. The first law of thermodynamics can be written as follows:

ΔQ = ΔU + W

ΔQ = mΔu + PΔV

where,

ΔQ = Heat transfer between system and surrounding = ?

Δu = specific change in internal energy of the system = - 175 KJ/kg

m = mass of air = 20 g = 0.02 kg

P = Constant Pressure = 101.3 KPa

ΔV = Change in Volume = 0.05 m³ - 0.005 m³ = 0.045 m³

Therefore,

ΔQ = (0.02 kg)(-175 KJ/kg) + (101.3 KPa)(0.045 m³)

ΔQ = -3.5 KJ + 4.56 KJ

4 0

2 years ago

5. Which statement regarding a finite state machine (FSM) is NOT true: (a) In a non-deterministic FSM, a string is invalid if th

valkas [14]

Answer:

The statement (a) In a non-deterministic FSM, a string is invalid if there is one path not leading to a final state is NOT true

Explanation:

A non-deterministic FSM, contrary to deterministic FSM which has only one possible thread of execution, has multiple threads and for the machine to be invalid, all threads should lead to a none accepting (final) state.

7 0

3 years ago