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BartSMP [9]
3 years ago
11

I need help with this answer

Physics
1 answer:
denpristay [2]3 years ago
6 0

Answer:

Single Replacement

Explanation:

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Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of
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Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) <em>High and low pressures in kilopascals</em>:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{high} = 15.999\,kPa

p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{low} = 10.666\,kPa

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) <em>High and low pressures in pounds per square inch</em>:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}

p_{high} = 2.320\,psi

p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}

p_{low} = 1.547\,psi

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) <em>High and low pressures in meter water column in meters water column</em>:

We can calculate the equivalent water column of a mercury column by the following relation:

\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}

h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg} (Eq. 1)

Where:

\rho_{w}, \rho_{Hg} - Densities of water and mercury, measured in kilograms per cubic meter.

h_{w}, h_{Hg} - Heights of water and mercury columns, measured in meters.

If we know that \rho_{w} = 1000\,\frac{kg}{m^{3}}, \rho_{Hg} = 13600\,\frac{kg}{m^{3}}, h_{Hg, high} = 0.120\,m and h_{Hg, low} = 0.080\,m, then we get that:

h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m

h_{w, high} = 1.632\,m

h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m

h_{w, low} = 1.088\,m

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

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WILL GIVE BRAINLIST: Does an objects weight affect how fast it falls?
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The pressure on a fluid at rest in a pipe increases by 20 Pa. How does this change in pressure affect the pressure on the fluid
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r^² x Pressure (pa).

Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.

Explanation:

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