Class l Help HW 8 Begin Date: 3/27/2018 11:59:00 AM -Due Date: 4/5/2018 11:59:00

Question

Class l Help HW 8 Begin Date: 3/27/2018 11:59:00 AM -Due Date: 4/5/2018 11:59:00 PM End Date: 5/12/2018 11:59:00 PM (6%) Problem 12. A certain ngid aluminum container contains a liquid at a gauge pressure off 0-2.02-10° Pa at sea level where the atmospheric pressure is P-1.01 x 105 Pa. The volume of the container is Vo-4.55 x 104m3. The maximum difference between the pressure inside and outside that this particular container can withstand before bursting or imploding is APmax 2.32 x 10% Pa. 1.20 kg /m and that the density of seawater For this problem, assume that the density of air maintains a constant value of pa maintains a constant value ofPs = 1025 kg / m3 25% Part (a) The container is taken from sea level, where the pressure ofair is P,-1.01x 105 Pa, to a higher altitude. What is the maximum height h in meters above the ground that the container can be lifted before bursting? Neglect the changes in temperature and acceleration due to gravity with altitude Grade Summary Deductions 0% Potential 100% sin0 coso tan0 cotan asin acoso Attempts remaining: 2 per attempt) detailed view tan) acotan sin0 cosh0 tanh cotanho ODegrees Radians Hints:-% deduction per hint. Hints remaining Feedback: 0% deduction per feedback. 25% Part (b) If we include the decrease in the density of the air with increasing altitude, what will happen? 25% Part (c) Choose the correct answer from the following options. ? 25% Part (d) What is the maximum depth d x in meters below the surface of the ocean that the container can be taken before imploding? Allcontent © 2018 Expert TA, LLC

Explanation / Answer

pressure inside the contained=P=atmospheric pressure+gauge pressure

=3.03*10^5 Pa

part a:

let at height h_max , the container explodes.

at h_max, atmospheric pressure=Pa-density of air*g*h_max

as the container explodes, the pressure outside should be P-2.32*10^5

=0.71*10^5 Pa

hence 1.01*10^5-1.2*9.8*h_max=0.71*10^5

==>h_max=2551 m

part b:

as density decreases, drop in pressure will be less as compared to density being constant.

in that case, the container will burst at a higher height.

hence the fifth option is correct.

“the height that the container could be lifted to without bursting would increase”

part c:

at maximum depth d_max, pressure outside required to burst the container should be P+2.32*10^5

=5.35*10^5 Pa

at depth d_max , pressure=Pa+density of water*g*d_max

=1.01*10^5+1000*9.8*d_max

hence 5.35*10^5=1.01*10^5+1000*9.8*d_max

==>d_max=44.286 m

hence first option is correct.

“the depth below the surface of the ocean the container can reach before exploding is less than the altitude in atmosphere the container can reach before imploding.”

part d:

d_max==44.286 m

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