Mga tamang paraan ng paggamit ng gamot

I need help starting this

Nesterboy [21]

Answer:

?

Explanation:whats the subject

4 0

3 years ago

Suppose a starbucks tall latte cost $4.00 in the united states, 5.00 euros in the euro area and $2.50 australian dollars in aust

Andrews [41]

Answer:B

Explanation:

5 0

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.

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

8 0

2 years ago

In classifying the kinds of projects an organization has in its portfolio, projects that are typically needed to support current

erastova [34]

Compliance (must do) projects.

6 0

2 years ago

Would you prefer a fully taxable investment earning 8.1 percent or a tax-exempt investment earning 6.1 percent? (assume a 28 per

blsea [12.9K]

<span>Prefer the 6.1 percent tax-exempt investment. Let's do the math and see why the tax-exempt investment is the better choice. For the 8.1% taxable investment, you get taxed at the rate of 28%. Which means that you only get to keep 100%-28% = 72% of your gains. So 0.72 * 8.1 = 5.832 which means your effective earning percentage is only 5.832% which is less than the 6.1% rate you get for the tax-exempt investment. Another consideration that wasn't taken into account for the question is the earnings on the taxable investment may push you up into a higher tax bracket. Which in turn increases the tax burden on your other investments. So the better choice here is the 6.1% tax-exempt investment even though that first glance the 8.1% investment looks higher.</span>

7 0

3 years ago

Mga tamang paraan ng paggamit ng gamot​

Mga tamang paraan ng paggamit ng gamot