Pleaseeee help me with this!!

A plastic molding machine produces a product whose annual demand is in the millions.

BabaBlast [244]

Answer:

a. 78.4 pieces/hr

b. $0.1806/min

c. $1.34/piece

Explanation:

(a) With a cycle time Tc = 45 sec = 0.75 min.

Production rate, Rp = 60/0.75 = 80 pieces/hr.

Let us factor in the 98% proportion uptime, so Production rate, Rp = 0.98*(80) = 78.4 pieces/hr

Quantity of product annualy = 6000 *(78.4) = 470,400 pieces/yr

(b) Equipment cost rate, Ceq = 500,000(1.30)/(60 x 10 x 6000) = $0.1806/min.

(c) Cost per piece of mould, Ct = 100,000/1,000,000 = $0.10/piece

Cost rate of labour ,CL = 18.00(0.20) = $3.60/hr = $0.06/min

Conclusively, cost per piece, Cpc = 1.20(0.88) + (0.06 + 0.1806)(0.75) + 0.10 = $1.34/piece

8 0

3 years ago

Helpppp EDO segundo orden

jeka94

Can’t really see it it’s blurry

Sorry

7 0

3 years ago

Select the correct answer.

bonufazy [111]

Answer:

OD.

plate

Explanation:

3 0

3 years ago

Seawater has a specific density of 1.025. What is its specific volume in m^3/kg (to 3 significant figures of accuracy, tolerance

Svetradugi [14.3K]

Answer:

$specific\ volume=0.00097\ m^3/kg$

Explanation:

Given that

Specific gravity of sea water = 1.025

So density of sea water = 1.025 x 1000 $kg/m^3$

Density of sea water = 1025 $kg/m^3$

We know that

$Density=\dfrac{mass}{Volume}$ ---1

Specific volume

$specific\ volume=\dfrac{Volume}{mass}$ ---2

From equation 1 and 2

We can say that

$specific\ volume=\dfrac{1}{density}\ m^3/kg$

$specific\ volume=\dfrac{1}{1025}\ m^3/kg$

$specific\ volume=0.00097\ m^3/kg$

6 0

3 years ago

An ideal gas is kept in a 10-liter [L] container at a pressure of 1.5 atmospheres [atm] and a temperature of 310 kelvins [k]. If

olchik [2.2K]

Answer: The new volume is 5 L.

Explanation: <u>Boyle's Law</u> describes the relationship between an ideal gas and pressure, volume and temperature, when temperature and the amount of gas are constant.

According to this law:

P₁V₁ = P₂V₂

Which states that, under those conditions, the pressure of the gas is inversely proportional to the volume.

It is also adequate to use this law when you want to determine pressure or volume at its initial or final value.

Since it is asked the final volume:

P₁V₁ = P₂V₂

V₂ = $\frac{P_{1}.V_{1} }{P_{2} }$

V₂ = $\frac{1.5.10}{3}$

V₂ = 5

When the gas is compressed until pressure of 3 atm, its volume is 5 liters.

4 0

3 years ago