Andy has applied for a NICET

Which items are NOT found on a

Alja [10]

Answer:

None of the above cause thats what i put

3 0

2 years ago

Explain the process of fractional distillation (include cracking)

Grace [21]

Answer:

Explanation:

Fractional distillation is the separation of a mixture into its component parts, or fractions. Chemical compounds are separated by heating them to a temperature at which one or more fractions of the mixture will vaporize. It uses distillation to fractionate

it is the separation of a liquid mixture into fractions differing in boiling point (and hence chemical composition) by means of distillation, typically using a fractionating column.

cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either: heated to 600-700°C.

I hope this was helpful

5 0

2 years ago

A thin flat plate of surface area, 500 cm², is located between two fixed plates such that the gap below the top plate is 20 mm a

kodGreya [7K]

Answer:0.3166N

Explanation:

Given data

Area $\left ( A\right )=500 cm^2$

Gap below top plate $\left ( y_1\right )=20 mm$

Gap above bottom plate $\left ( y_2\right )=30 mm$

SAE 30 oil viscosity = $0.38 N-s/m^2$

Velocity of middle plate $\left ( v\right )=1 m/s$

There will viscous force on middle plate i.e. at above surface and below surface

Viscous force $\left ( F\right )=\mu \frac{Av}{y}$

$Net Force on plate F =\mu Av\left [\frac{1}{y_1} +\frac{1}{y_2}\right ]$

$F=0.38\times 500\times 10^{-4}\left [\frac{1}{20\times 10^{-3}} +\frac{1}{30\times 10^{-3}}\right ]$

$F=31.66\times 10^{-2}=0.3166 N$

this is force by oil on plate thus we need to apply atleast 0.3166N to move plate

8 0

3 years ago

. For each of the following situations, determine if the diode below is in foward or reversebias(a) (2 points) V1 = 0 V, V2= 2V(

Mazyrski [523]

Assuming V1 is the anode and v2 the cathode (Anode is P region and Cathode is N)

Answer:

a) Reverse bias

b) Forward bias

c) Forward bias

Explanation:

Forward bias: It happens whenever the N region of the diode is more positive than the P region. Hence, the depletion zone increase ceasing the current through the circuit -> V1 -V2 < 0

Reverse bias: It happens whenever the P region of the diode is more positive than the N region. In this case, the depletion zone begins to shrink, if enough voltage is applied current could go through the circuit -> V1 - V2 > 0

a) V = V1 - V2 = 0 - 2 = -2 -> -2 is smaller than zero therefore, we have reverse bias

b) V = V1 - V2 = 4.5 - 2.8 = 1.7 -> 1.7 is greater than zero therefore, we have forward bias

c9 V = V1 - V2 = -1 - -1.3 = 0.3 -> 0.3 is greater than zero therefore, we have forward bias

6 0

3 years ago

Water flows at a rate of 0.040 m3 /s in a horizontal pipe whose diameter is reduced from 15 cm to 8 cm by a reducer. If the pres

Blababa [14]

Answer:

hL = 0.9627 m

Explanation:

Given

Q = 0.040 m³/s (constant value)

D₁ = 15 cm = 0.15 m ⇒ R₁ = D₁/2 = 0.15 m/2 = 0.075 m

D₂ = 8 cm = 0.08 m ⇒ R₂ = D₂/2 = 0.08 m/2 = 0.04 m

P₁ = 480 kPa = 480*10³Pa

P₂ = 440 kPa = 440*10³Pa

α = 1.05

ρ = 1000 Kg/m³

g = 9.81 m/s²

h₁ = h₂

hL = ? (the irreversible head loss in the reducer)

Using the formula Q = v*A ⇒ v = Q/A

we can find the velocities v₁ and v₂ as follows

v₁ = Q/A₁ = Q/(π*R₁²) = (0.040 m³/s)/(π*(0.075 m)²) = 2.2635 m/s

v₂ = Q/A₂ = Q/(π*R₂²) = (0.040 m³/s)/(π*(0.04 m)²) = 7.9577 m/s

Then we apply the Bernoulli law (for an incompressible flow)

(P₂/(ρ*g)) + (α*v₂²/(2*g)) + h₂ = (P₁/(ρ*g)) + (α*v₁²/(2*g)) + h₁ - hL

Since h₁ = h₂ we obtain

(P₂/(ρ*g)) + (α*v₂²/(2*g)) = (P₁/(ρ*g)) + (α*v₁²/(2*g)) - hL

⇒ hL = ((P₁-P₂)/(ρ*g)) + (α/(2*g))*(v₁²-v₂²)

⇒ hL = ((480*10³Pa-440*10³Pa)/(1000 Kg/m³*9.81 m/s²)) + (1.05/(2*9.81 m/s²))*((2.2635 m/s)²-(7.9577 m/s)²)

⇒ hL = 0.9627 m

7 0

3 years ago