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3 years ago

Products that are in the process of being manufactured but are not yet complete are called:

Engineering

1 answer:

vazorg [7]3 years ago

6 0

Answer:

Those products are generally called Work in Process WIP

Explanation:

Work in process (WIP), or work in progress (WIP), goods in process, or in-process inventory in a manufacturing industry/company refer to the company's partially finished goods waiting for completion and eventual sale or the value of these items.

These items are either just being produced or require further processing (like purification, separation, packaging or handling) in a queue or a buffer storage.

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yolanda went to get her car fixed. it cost $50 for the part and $200 for each hour the mechanic worked on it. if x represents th

sergij07 [2.7K]

The amount that a dealership pays a manufacturer for a car is essentially its invoice price.

A car with a $28,000 MSRP, for instance, can have an invoice price of about $26,000, which means that the dealer paid the manufacturer $26,000 for the car. Repair costs do not include operating costs and refer to costs and expenses that are reasonably necessary or desirable owing to routine wear and tear, periodic replacement, vandalism, damage to, or destruction of any part of the facility property. In the simplest terms, total cost (TC) is the sum of all expenses spent over the course of production or activity. Total cost in economics is the sum of variable and fixed costs.

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1 year ago

A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125

Gekata [30.6K]

Answer:

a) $\dot m_{3} = 135\,\frac{lbm}{s}$ , b) $h_{3}=168.965\,\frac{BTU}{lbm}$ , c) $T = 200.829\,^{\textdegree}F$

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

$\dot m_{1} + \dot m_{2} - \dot m_{3} = 0$

The mass flow rate exiting the tank is:

$\dot m_{3} = \dot m_{1} + \dot m_{2}$

$\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}$

$\dot m_{3} = 135\,\frac{lbm}{s}$

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

$\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0$

$h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}$

Properties of water are obtained from tables:

$h_{1}=180.16\,\frac{BTU}{lbm}$

$h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)$

$h_{2}=29.032\,\frac{BTU}{lbm}$

The specific enthalpy at outlet is:

$h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }$

$h_{3}=168.965\,\frac{BTU}{lbm}$

c) After a quick interpolation from data availables on water tables, the final temperature is:

$T = 200.829\,^{\textdegree}F$

8 0

3 years ago

Read 2 more answers

What is an air mass?

kotegsom [21]

Answer:

An air mass is a body of air with horizontally uniform temperature, humidity, and pressure.

Explanation:

Because it is

8 0

3 years ago

Read 2 more answers

Block D of the mechanism is confined to move within the slot of member CB. Link AD is rotating at a constant rate of ωAD = 6 rad

svet-max [94.6K]

Answer:

1) 1.71 rad/s

2) -6.22 rad/s²

Explanation:

Choose point C to be the origin.

Using geometry, we can show that the coordinates of point A are:

(a cos 30°, a sin 30° − b)

Therefore, the coordinates of point D at time t are:

(a cos 30° − b sin(ωt), a sin 30° − b + b cos(ωt))

The angle formed by CB with the x-axis is therefore:

tan θ = (a sin 30° − b + b cos(ωt)) / (a cos 30° − b sin(ωt))

1) Taking the derivative with respect to time, we can find the angular velocity:

sec² θ dθ/dt = [(a cos 30° − b sin(ωt)) (-bω sin(ωt)) − (a sin 30° − b + b cos(ωt)) (-bω cos(ωt))] / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω [(a cos 30° − b sin(ωt)) sin(ωt) − (a sin 30° − b + b cos(ωt)) cos(ωt)] / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω [(a cos 30° sin(ωt) − b sin²(ωt)) − (a sin 30° cos(ωt) − b + b cos²(ωt))] / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − b sin²(ωt) − a sin 30° cos(ωt) + b − b cos²(ωt)) / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -bω (a cos 30° sin(ωt) − a sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²

sec² θ dθ/dt = -abω (cos 30° sin(ωt) − sin 30° cos(ωt)) / (a cos 30° − b sin(ωt))²

We know at the moment shown, a = 350 mm, b = 200 mm, θ = 30°, ω = 6 rad/s, and t = 0 s.

sec² 30° dθ/dt = -(350) (200) (6) (cos 30° sin(0) − sin 30° cos(0)) / (350 cos 30° − 200 sin(0))²

sec² 30° dθ/dt = -(350) (200) (6) (-sin 30°) / (350 cos 30°)²

dθ/dt = (200) (6) (1/2) / 350

dθ/dt = 600 / 350

dθ/dt = 1.71 rad/s

2) Taking the second derivative of θ with respect to time, we can find the angular acceleration.

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30° − b sin(ωt))² (ω cos 30° cos(ωt) + ω sin 30° sin(ωt)) − (cos 30° sin(ωt) − sin 30° cos(ωt)) (2 (a cos 30° − b sin(ωt)) (-bω cos(ωt)))] / (a cos 30° − b sin(ωt))⁴

At t = 0:

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω [(a cos 30°)² (ω cos 30°) − (0 − sin 30°) (2 (a cos 30°) (-bω))] / (a cos 30°)⁴

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -abω (a²ω cos³ 30° − 2abω sin 30° cos 30°) / (a⁴ cos⁴ 30°)

sec² θ d²θ/dt² + 2 sec² θ tan θ dθ/dt = -bω (aω cos² 30° − 2bω sin 30°) / (a² cos³ 30°)

d²θ/dt² + 2 tan θ dθ/dt = -bω² (a cos² 30° − b) / (a² cos 30°)

Plugging in values:

d²θ/dt² + 2 tan 30° dθ/dt = -(200) (6)² (350 cos² 30° − 200) / (350² cos 30°)

d²θ/dt² + 2 tan 30° dθ/dt = -7200 (262.5 − 200) / (350² cos 30°)

d²θ/dt² + 2 tan 30° (1.71) = -4.24

d²θ/dt² = -6.22 rad/s²