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4 years ago

What is the differences between stack and queue?

Engineering

1 answer:

Montano1993 [528]4 years ago

7 0

Answer:

A stack is an ordered list of elements where all insertions and deletions are made at the same end, whereas a queue is exactly the opposite of a stack.

Explanation:

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You must signal _____ before any turn or lane change. A. 5 seconds B. 10 seconds C. 50 ft D. 100 ft

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The answer is D! hope you pass

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3 years ago

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What would happen if an exposed film was accidentally placed in the fixer before being placed in the developer

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3 years ago

For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh

Paul [167]

Answer:

the pressure drop is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere $dp$ = 0.003 m

particle density = 1300 kg/m³

the bed void fraction $\in =$ 0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas $\rho$ = 0.15 mol/dm ⁻³

viscosity of methane gas $\mu$ = 1.429 x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³ to kg/m³

SO; we have :

Density = 0.15 mol/dm ⁻³

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density = $0.1 5 *\dfrac{16}{0.1^3}$

Density = 2400

Density $\rho_f$ = 2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas = 4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

$Re = \dfrac{dV \rho}{\mu}$

$Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}$

$Re=2276.317705$

For Re > 1000

$\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}$

$\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}$

$\Delta P=8575.755212*2.5$

$\Delta = 21439.38803 \ Pa$

To atm ; we have

$\Delta P = \dfrac{21439.38803 }{101325}$

$\Delta P =0.2115903087 \ atm$

ΔP ≅ 0.21159 atm

Thus; the pressure drop is 0.21159 atm

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3 years ago

A manufacturing company has decided to be carbon free by 2030.

k0ka [10]

Answer:

company should focus on their goals

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3 years ago

An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.0205 Ω, and a 1.53-V carbon-z

mina [271]

Answer:

(a) The current in amperes that flows through the toy's circuit is 0.923A

(b) The power supplied to the toy is 5.78721W

(c) The internal resistance r2 of the failed dry cell is 72Ω

Explanation:

From the circuit diagram attached. We have the electric component:

B1 = 3* 1.58 = 4.74V

B2 = 1.53V

r1 = 3*0.0205 = 0.0615Ω

r2 = 0.105Ω

R = 6.625Ω

Since the internal resistances and the resistor R are connected in series, we can calculate the total resistance RT as

RT = r1 + r2 +R = 0.0615 + 0.105 + 6.625

= 6.7915Ω

Total Voltage supplied to the circuit by both batteries V = B1 + B2 = 4.74 + 1.53 = 6.27V

(a) CIRCUIT CURRENT

The current I, flowing through the circuit is i = $\frac{V}{R_{T}}$ = $\frac{6.27}{6.7915}$ =0.923A

The current in amperes that flows through the toy's circuit is 0.923A

(b) THE POWER SUPPLIED TO THE TOY

Power P = I*V =0.923*6.27 = 5.78721W

The power supplied to the toy is 5.78721W

Due to dry cell failure, the power supplied to the toy is reduced to 0.5W

Now Power P = $\frac{V^{2} }{R}$ . To calculate the new total resistance of the circuit we will make R the subject of the formular

R= $\frac{V^{2} }{P}$ = $\frac{6.27^{2} }{0.5}$ = $\frac{39.3129}{0.5}$ = 78.6258Ω

Remember that RT = r1 + r2 + R

r1 =RT- (R +r2)

r1 = 78.6258 - 6.6865 =71.9393Ω

The internal resistance r2 of the failed dry cell is 72Ω

6 0

3 years ago

Read 2 more answers