<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r
v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span>
<span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span>
<span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min).
(32.68 rev/s)(60 s/min)=<span>
<span>1960.74 rev/min
</span></span>
Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
To be referenced, it would be true
Explanation:
answer is attached, kindly see answer
