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sergeinik [125]
2 years ago
12

When you hold your hands at your sides, you may have noticed that the veins sometimes bulge--the height difference between your

heart and your hands produces increased pressure in the veins. The same thing happens in the arteries. If the average arterial pressure at your heart is a typical 100 mmHg, what is the average arterial pressure in your hands when they are held at your side? Assume your hands are 60 cm below your heart.
Physics
1 answer:
hodyreva [135]2 years ago
7 0

To solve this problem it is necessary to use the concepts related to pressure and pressure, absolute and atmospheric.

Average arterial pressure in the hands,

P = 100mmHg+ h*\rho_{blood}g

Where,

P = Pressure

h = height (at this case the length of the arm)

Replacing with our values

P = 100mm(Hg)+(600mm)(\frac{\rho_{blood}}{\rho_{mercury}})(Hg)

P = 100mmHg+600*\frac{1060}{13600}mmHg

P = 100mmHg+46.765mmHg

P = 146.765mmHg

Where,

\rho_{blood} = 1060Kg/m^3

\rho_{mercury}=13600Kg/m^3

Therefore the pressure is 146.765mmHg

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Anything that's not supported and doesn't hit anything, and
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                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

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7 0
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If it were possible to remove gravity and friction, think about what would happen to a football if it were tossed into the air.
elena-14-01-66 [18.8K]
Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>
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Cathode ray tubes in old television sets worked by accelerating electrons and then deflecting them with magnetic fields onto a p
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B = 0.046T

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2 years ago
A car drives toward the right over the top of a hill, as shown below. An illustration of car at the top of a hill pointing right
Sav [38]

Answer: X

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A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

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2 years ago
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