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andriy [413]
3 years ago
10

Dan has been directed to study the forces close to a company that affect its ability to serve its customers, such as the company

, suppliers, marketing intermediaries, customer markets, competitors, and publics. In this instance, Dan has been directed to study the ________ of the company.
Business
1 answer:
shutvik [7]3 years ago
5 0

Answer:

Micro environment                                              

Explanation:

The institution's micro-environment comprises of those components that are manageable. Generally the micro-environment doesn't really impact all businesses in a sector in same manner, as the scale, efficiency, competence and approaches are different.

For instance, the suppliers of raw materials are giving big corporations more compromises. We might not offer small businesses the same concessions though. Thus, from the above we can conclude that the correct answer is micro-environment.

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Maroon hotels inc. is using the behaviorally anchored rating scale (bars) method of employee performance appraisal. what is the
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Adventure Holidays sells thousands of tour packages each month through its branches. A branch manager's salary would be a(n) ___
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Indirect cost

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3 years ago
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

 

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