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2 years ago

Describe each occupation

Engineering

1 answer:

vitfil [10]2 years ago

5 0

Answer:

Carpenter — A carpenter is someone who works with wood. They build houses and make cabinets etc.

Logging — Loggers are people who cut trees. They use strong chain saws to cut trees.

Shipwrights — Shipwrights build, design, and repair all sizes of boats.

Wood Machinist — Wood Machinists repair and cut timber or any kind of wood for construction projects. They also operate woodworking machines, as their name suggest.

Furniture finishers — Furniture finishers shape, decorate, and restore damaged and worn out furniture.

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A round bar of chromium steel, (ρ= 7833 kg/m, k =48.9 W/m-K, c =0.115 KJ/kg-K, α=3.91 ×10^-6 m^2/s) emerges from a heat treatmen

Lerok [7]

Answer:

Q = 424523.22 kw

Explanation:

$\rho =7833 kg/m$

k = 48.9 W/m - K

c = 0.115 KJ/kg- K

$\alpha = 3.91*10^{-6} m^2/s$

$T_s = 285 degree celcius$

T_∞ = 35 degree celcius

velocity of air stream = 15 m/s

D = 40 cm

L = 200 cm

mass flow rate $\dot m = \rho AV = 7833 *\frac{\pi}{4} 0.4^2*15$

$\dot m = 14764.85 kg/s$

$A_s = \pi DL = \pi 0.4*2 = 2.513 m^2$

$Q = \dot m C \Delta T = h A_s \Delta T$

$\dot m C \Delta T = h A_s \Delta T$

solving for h

$h = \frac{14764.85*0.115*(285-35)}{2.513*(285-35)}$

h = 675.6 kw/m^2K

$Q = h A_s\Delta T$

Q = 675.6*2.513*(285-35)

Q = 424523.22 kw

7 0

2 years ago

A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t

Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0

3 years ago

You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the

defon

You need to explain it more simple as everyone is clueless

7 0

2 years ago

We can process oil into a lot of useful fuels to run our cars, trucks, and even airplanes. Oil is used for making lots of other

Ostrovityanka [42]

Answer:

Explanation:

Products of oil in our everyday life:

(1) Petro-Chemical Feedstock: These are by product of Refining of Oil which it is used extensively to make PET bottles, Paints, Polyester Shirts, Pocket combs e.t.c

(2) Asphalt : Used extensively to make Motor Road, highways

(3) Plastics : we use plastics in our everyday life, this is also a product of Refining of crude oil e.g PVC, Telephone casing, Tapes e.t.c

(4) Lubricating Oil/Grease : This is another product from crude oil Fractional Distillation.

(5) Propane/ Cooking Gas: This is also a product from oil which is used in our everyday life for cooking, grilling etc.

4 0

3 years ago

The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u

zimovet [89]

Answer:

F = 0.0022N

Explanation:

Given:

Surface area (A) = 4,000mm² = 0.004m²

Viscosity = µ = 0.55 N.s/m²

u = (5y-0.5y²) mm/s

Assume y = 4

Computation:

F/A = µ(du/dy)

F = µA(du/dy)

F = µA[(d/dy)(5y-0.5y²)]

F = (0.55)(0.004)[(5-1(4))]

F = 0.0022N

8 0

3 years ago