ch10 - Physics 101 W Abdul-Ramq Chapter/0 Homework Solution...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 101 ' W. Abdul-Ramq Chapter /0 Homework Solution 2. mafV: 1-27 (wixmyz'm = {Z (abbot/oi W W55 2’ 9‘ 1 mm-Hg J g. AP:pm=(Losstkg/m=)(9.som/sz)<x-wm>=1-646x10‘N/m’fl‘ssN/m2 9. (a) The total force of the atmosphere on the table will be the air pressure times the area of the table. ! F=PA=(1.013><10’ N/m’)(1.6 m)(2.9 m): (b) Since the atmospheric pressure is the same on the underside of the table (the height difference is minimal), the upward force of air pressure is thesame as the downward force of air on the top of the table, . V ' 11. The sum of the force exerted by the pressure in each is equal to the weight of the car. 4(2.40xlo’ N/m2)(2200m2)[11m2> " 4m mg=4PA —-) m=T= (glsom/sz) 14. (a) The absolute pressure is given by Eq. 10-3c, and the total force is the absolute pressure times the area of the bottom of the pool. 13:}; +pgh=1.013x10’ N/m2 .+(1.00x103kg/m’)(9.som/s’)(2.o m) = F = PA =(1.21x10’ N/m’)(22.o m)(8.5 m) = (b) The pressure against the side of the pool, near the bottom, will be the same as the pressure at the bottom, P = 1.21xlo’ N/m2 ' 43. We assume that there is no appreciable height difference betweenthe two sides of the roof. Then the ’ net force on the roof due to the air is the difference in pressure on the two sides of the roof, times the area of the roof. The difference in pressurecan be found from Bemoulli’s equation. E The lift force would be the difference in pressurebetween the two wing surfaces, times the area of the wing surface. The difference in pressure can be found from Bernoulli’s equation. We consider the two surfaces of the wing to be at the same height above the ground. Call the bottom surface of the wing point 1, and the top surface point 2. R +ipv? +Pey. = P2 “mi +sz _-* R-Pz =’;P(v§ rvf) FM = (PI — P,)(Area of wing) =-;-p(v; —v,2)A = {(1.29 iqg/m’)[(260m/s)2 -(150 m/s)’](7sm‘) = 45. . The air pressure inside the hurricane can be estimated using Bernoulli’s equation. Assume the pressure outside the hurricane is air pressure, the speed of the wind outside thehurricane is 0, and that the two pressure measurements are made at the same height. Pinside +‘i'pvi2mide + pgyimide = Poutside +§Pvfim + ngm ") ._ ...L 2 Puma " P outside 2 plirvimide 1000m 1h 2 =1.013xio‘pa—g(1.29kg/m’)[(300km/h)(—k:n—)[gmfl = a 0.96 atm @ Use the equation of continuity (Eq. 10-4) to relate the volume flow of water at the two locations, and use Bernoulli's equation (Eq. 10-5) to relate the conditions at the street to those at the top floor. ' Express the pressures as atmospheric pressure plus gauge pressure. Amuvmet = vam -> 2 V -2 v =vmi“!-=(o.60nt/s)”(5'0>(10 m): =2.219m/s~ ‘°" A... zz(2.6x10'2 m) Po +P..... +ipvf... +pgyw = R. +ng... +ipvi, +pgym, -» street Pp... =P +ip(vfm -vip)+P8y(ym-yw) lull: top meet = (3.8atm)[L9_1_3_1(_1—Qs—1:9-]+-;-(1.00x103 kg/m3)[(0.60m/s)2 —(2.219m/s)2]2 atm é g +(1.oox103kg/m’)(9.8m/s2)(—18m) 1: 2'0 “‘9 Pa r... . -....—.~ 3 ...
View Full Document

This note was uploaded on 10/13/2010 for the course PHYSICS 101 taught by Professor Abdulrazzaq during the Spring '10 term at West Virginia State University.

Page1 / 3

ch10 - Physics 101 W Abdul-Ramq Chapter/0 Homework Solution...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online