Question

Text 1, 2, 3, 5, 8,11, 14, 20, 22 Water 1. An open-ended can 1 ft long is originally full of air at 70°F. The can is now immersed in water. Assuming that the air stays at 70°F and behaves as an ideal gas, how high will the water rise in the can? i0 ft I ft Air 2. Let's take a look at the images. There is an a)

how did you calculate pressure of p(air) ? the blue one that i wrote.

i dont understand to write in cursive script ,, sorry

Answer 1

The immersed can is shown in the following figure: Air In the figure, h is the depth of immersion to the bottom of the tank, h, is the height of the can, and ha height to which water is filled in the can. The pressure of air is as follows: Por = p m ) Pair a h-h Here, P, is the atmospheric pressure and the value of the atmospheric pressure is Pa = 14.7 lbf/in The expression for the pressure of the water is as follows: Pwater = Pa+pg(h-h) Here, p is the density of water and the value of density of water is p = 62.4 lbs/ft'.

inside the can is balanced by the pressure of water At equilibrium, the pressure of air P Pwater inside the can. That is, Pa+pg(h Pair = Pirater Pe+P8(–h,) = P. (main) pg (=h)= P. A. 1) PE(4-1)=P. *) (hh)(h, - h) - Polo man - Wah – Woha + ing? - Pink mo - na pasta – na ]thim = 0

From the given data, the height of the water raised in can is as follows: hP-n pa man thus = 0 4-h +hh = 0 ( 32.2 lbs (144 in? 14.7 lbf/in2 1 lbf 1 ft2 ² - h 2-10 ft-1 ft +(10 ft)(1 ft) = 0 (62.4 lbs/ft)(32.2 ft/s?) h?? - (22.93)h +10 = 0 Solve the quadratic equation h? +(22.93 ft) h+(10 ft)=0 for hz. -(-22.93) + (-22.93)² - 4(10)(1) 2(1) = 0.44 ft] (or) 22.93 ft h = From the above values, 22.93 ft is impossible to raise the water because the height of the can is 1 ft. Hence, the height of the water raise in can is 0.44 ft.