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2 years ago

How deep is a 6ft hole?

Engineering

2 answers:

Pavlova-9 [17]2 years ago

3 0

Answer:

I know this sounds quite deep but it is as deep as a grave

Explanation:

It's reality

denis23 [38]2 years ago

3 0

Answer:

6ft or 72 inches

Explanation:

The hole is most likely 6ft or also known as 72 inches.

Hope this helps :-)

Have a great rest of your day or night!

<3 simplysun

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Air flows from a large reservoir in which the pressure and temperature are 1 MPa and 30°C, respectively, through a convergent–di

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A rigid tank contains 2 kg of N2 and 4 kg of Co2 at temperature of 25 C and 1 MPa. Find the partial pressure of each gas respect

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Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.

Explanation: Dalton's Law of Partial Pressure states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:

$P_{total} = P_{1}+P_{2}+...$

The proportion of each individual gas in the total pressure is expressed in terms of mole fraction:

$X_{i}$ = moles of a gas / total number moles of gas

The rigid tank has total pressure of 1MPa.

Nitrogen gas:

molar mass = 14g/mol

mass in the tank = 2000g

number of moles in the tank: $n=\frac{2000}{14}$ = 142.85mols

Carbon Dioxide:

molar mass = 44g/mol

mass in the tank = 4000g

number of moles in the tank: $n=\frac{4000}{44}$ = 90.91mols

Total number of moles: 142.85 + 90.91 = 233.76 mols

To calculate partial pressure:

$P_{i}=P_{total}.X_{i}$

For Nitrogen gas:

$P_{N_{2}}=1.\frac{142.85}{233.76}$

$P_{N_{2}}$ = 0.6

For Carbon Dioxide:

$P_{total}=P_{N_{2}}+P_{CO_{2}}$

$P_{CO_{2}} = P_{total}-P_{N_{2}}$

$P_{CO_{2}}=1-0.6$

$P_{CO_{2}}=$ 0.4

Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.

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2 years ago

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

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Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

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3 years ago

What is the inductive reactance of a 20 mH inductor at a frequency of 100Hz?

Inessa05 [86]

Answer:

The correct approach is "12.56 Ω".

Explanation:

The given values are:

Frequency,

f = 100 Hz

Inductor length,

L = 20 mH

Now,

The inductive reactance will be:

⇒ $X_L=2 \pi fL$

On putting the estimated values, we get

⇒ $=2 (3.14)\times 100\times 20\times 10^{-3}$

⇒ $=12.56 \ \Omega$

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3 years ago

How deep is a 6ft hole?​

How deep is a 6ft hole?