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vodomira [7]
3 years ago
9

A manufacturer reports the following costs to produce 10,000 units in its first year of operations:

Business
1 answer:
rewona [7]3 years ago
7 0

Answer:

Option (C) is correct.

Explanation:

Variable overhead per unit:

= Variable overhead ÷ Total units produced

= $70,000 ÷ 10,000

= $7 per unit

Fixed overhead per unit:

= Fixed overhead ÷ Total units produced

= 120,000 ÷ 10,000

= $12 per unit

Total product cost:

= Direct materials + Direct labor + Variable overhead + Fixed overhead

= 10 + 6 + 7 + 12

= $35 per unit

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When you get a phishing email, your response should be as follows:

B. Check for phishing by hovering over the hyperlink, checking the sender's name, and reviewing for improper grammar and misspelled words.

E) Click the "Report Phish" button to report the suspicious email.

<h3>What is email phishing?</h3>

Email phishing is the sending of fraudulent messages to trick recipients into revealing sensitive information by making them think that they had become extremely lucky without participating in any lottery or game.

When email phishing is received, do not click the link or forward the email to another person.  You do not need to respond to the sender, asking for its legitimacy.

2. The best way to navigate through the knowledge CHECK is <u>A. Click the Back button</u> to return to the video.

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4 0
2 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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The answer is shoring
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~

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