Please answer ALL parts of the question found on both photos(Part A through H).

Question

Please answer ALL parts of the question found on both photos(Part A through H). Show all work without skipping any steps. Use an equation editor or a pen and paper only. Thanks so much!

PlI 202 Winter 2018 Mesigment Sar Tool Physics in Nursing and Cardiology (Default) March 2018 nns pectain to the situation described below iven Density of mereury- 13.6 x 10 kg atm-1.013 x10 Pa solue puts a-h A mercury barometer is shown below. A column of mercury is supported by atmospheric pressure pusang the base. The space at the top of the glass tube is evacuated to vacuum pressare. How tall of a column ot mercury does standard atmospheric pressure support How all or a Nnrealmn des The barometer horn the previous qurton s now filed with water. lfaperson's blood pressure is12smrnlig oc85mmW. what is their maannal peaae, mmHg UI Forn mean arterial preseure of 100 mmHg·celeuine the puge pressure" P. prodied by the ban.

Explanation / Answer

a. since the poriont of space inside the tube above mercury is vaccum
let eheight of the mercury column be h
then
density of mercuty, rho = 13.6*10^3 kg/m^3
hence
rho*g*h = 1.01*10^5 Pa
h = 1.01*10^5/9.81*13.6*10^3 = 0.757030641 m column of mercury

b. when water is filled instead of mercury
rho = 1000 kg/m^3
hence
h = 1.01*10^5/9.81*1000 = 10.295616717 m column of water

c. given P1 = 125 mmHg
P2 = 85 mm Hg
mean pressure P = (P1 + P2)/2 = 105mm Hg
in pascals
P = 0.105*1.01*10^5/0.757030641 = 1.40220216*10^4 Pa

d. 100 mm Hg will produce a gauge pressure of 0.1*1.01*10^5/0.757030641
dP = 1.33416*10^4 Pa

e. h = 1.3 m
hb = 1.75 m
gauge pressure at heart = 1.33416*10^4 Pa
hence
gauge pressure at feet = 1.33416*10^4 + 1050*9.81*(hh) = 2.673225*10^4 Pa

f. gauge pressure at braing height = 1.33416*10^4 - 1050*9.81*(hb - h) = 8706.375 Pa
g. height from the floor saline solution will rise = dH
dH = 2.673225*10^4/1050*9.81 = 2.595238 m
h. if mean pressure at heart is 40 mm Hg, then the blood will not be able to reach the head of the person of this heigt

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