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MakcuM [25]
2 years ago
8

Krisp Bite, a breakfast cereal manufacturer, has a storage facility to store sugar, corn, wheat, oats, rice, and barley before t

hey are processed into cereal. In the context of operations management, Krisp Bite uses the storage facility to store its
A) throughput
B) information systems
C) contingency systems
D) inventory
Business
1 answer:
SIZIF [17.4K]2 years ago
4 0

Answer:

D) inventory

Explanation:

It is called inventory,( as the sugar , corn wheat , oats, rice and barley are direct materials not processed they are termed as inventory).

Choice D is the correct answer .

Choice A is not correct because throughput means output of something after processing .

Information system means systems relating to or showing information.

The sugar , corn wheat , oats, rice and barley are the raw direct materials used in the production of the cereals.

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Read 2 more answers
Pressure with Two Liquids, Hg and Water. An open test tube at 293 K is filled at the bottom with 12.1 cm of Hg, and 5.6 cm of wa
Radda [10]

Answer:

1170839.28 dyn/cm^2

16.9816 psia

117.083928 kN/m^2

Explanation:

To calculate the absolute pressure in the bottom of tube we need to sum the atmosferic and gauge pressure.

P_{abs}=P_{atm}+P_G

And the gauge pressure is given by the contributions of columns of water (P_{w}) and mercury(P_{Hg}), we can calculate the contribution of each column as:

P= \rho g h (*)

where \rho is the respective density, g gravity and h is height.

So we have all the data required to use the above equations (P_{atm}, height and density of each column) we only need to be carefully with the units.

For simplicity we can to express all pressure contributions in mmHg ( P_{atm}, P_{w} and P_{Hg}). Note that the units "x" mmHg  means the pressure at the bottom of a column of mercury of "x" mm high. For example, in this case we have a column 12.1 cm of Hg, that is a column of 121 mmHg (passing from cm to mm only requires multiply by 10) pressure exerted by that column is 121 mmHg.

Now pressure of 5.6 cm (56 mm) of water would be 56 mm of water, but it is not the same that mmHg, since the density of water is lower, the pressure exerted by 1 mm of water is lower than the exerted by 1 mm of Hg. The conversion between mmHg and mm of water is given by the relation between the densities.

mmHg=\frac{\rho_w*mmH_2O}{\rho_{Hg}}

mmHg=\frac{0.998*mmH_2O}{13.55}=0.0737 mmH_2O

And pressure of water in mmHg is

0.0737*56=4.1246 mmHg

The absolute pressure is:

P_{abs}=P_{atm}+P_G= 756 + 121 + 4.1246  = 881.1246 mmHg = 88.11246cmHg

To pass to dyn/cm^2 units we need to use the equation (*)

P= \rho g h = 13.55 \frac{g}{cm^3} * 980.665 \frac{cm}{s^2} * 88.11246 cmHg = 1170839.28 \frac{g}{cm s^2} = 1170839.28 \frac{dyn}{cm^2}

Note: We need to use cm Hg for units coherence

Now passing from dyn/cm^2 to kN/m^2 (or kPa) we need to consider that 1 dyn is 10^{-8} kN and 1 cm^2 is 10^{-4} m^2.  

1170839.28 \frac{dyn}{cm^2} * \frac{10^{-8}kN}{1 dyn}*\frac{cm^2}{10^{-4}m^2}=117.083928kN/m^2

Now passing kN/m^2 to psia. We need to consider that 1 psia is 6.89476.

117.083928kN/m^2*\frac{1psia}{6.89476kN/m^2}=16.9816 psia

 

8 0
3 years ago
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