Aumento da pressão

Livro halliday

Cap 14

Exercicio 1

1. The pressure increase is the applied force divided by the area: Δp = F/A = F/πr2, where r is the radius of the piston. Thus

Δp = (42 N)/π(0.011 m)2 = 1.1 × 105 Pa. This is equivalent to 1.1 atm.

2. We note that the container is cylindrical, the important aspect of this being that it has a uniform cross-section (as viewed from above); this allows us to relate the pressure at the bottom simply to the total weight of the liquids. Using the fact that 1L = 1000 cm3, we find the weight of the first liquid to be

W =mg=ρVg=(2.6g/cm3)(0.50L)(1000cm3 /L)(980cm/s2)=1.27×106g⋅cm/s2 1111

=12.7 N.

In the last step, we have converted grams to kilograms and centimeters to meters.

Similarly, for the second and the third liquids, we have

W =mg=ρVg=(1.0g/cm3)(0.25L)(1000cm3 L)(980cms2)=2.5N 2222

and

The total force on the bottom of the container is therefore F = W1 + W2 + W3 = 18 N.

3. The air inside pushes outward with a force given by piA, where pi is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by poA, where po is the pressure outside. The magnitude of the net force is F = (pi – po)A. Since 1 atm = 1.013 × 105 Pa,

F=(1.0atm−0.96atm)(1.013×105 Pa/atm)(3.4m)(2.1m)=2.9×104 N.

4. Knowing the standard air pressure value in several units allows us to set up a variety of conversion factors:

(80 mmHg) \0 1.01×105 Pa \0 = 10.6 kPa.

(a) P = 28 lb/in.2 \01.01×105 Pa \0 =190 kPa . ( )\014.7lb/in2 \0

\0\0

(b) (120 mmHg) \0 1.01×105 Pa \0 = 15.9 kPa,

\0760mmHg\0

\0\0 \0\0

5. Let the volume of the expanded air sacs be Va and that of the fish with its air sacs collapsed be V. Then

ρfish = mfish =1.08 g/cm3 and ρw = mfish =1.00 g/cm3 V V +Va

where ρw is the density of the water. This implies

ρfishV = ρw(V + Va) or (V + Va)/V = 1.08/1.00,

which gives Va/(V + Va) = 0.074 = 7.4%.

6. The magnitude F of the force required to pull the lid off is F = (po – pi)A, where po is the pressure outside the box, pi is the pressure inside, and A is the area of the lid. Recalling that 1N/m2 = 1 Pa, we obtain

p =p −F=1.0×105 Pa− 480N =3.8×104 Pa. i o A 77×10−4 m2

7. (a) The pressure difference results in forces applied as shown in the figure. We consider a team of horses pulling to the right. To pull the sphere apart, the team must exert a force at least as great as the horizontal component of the total force determined by “summing” (actually, integrating) these force vectors.

We consider a force vector at angle θ. Its leftward component is Δp cos θdA, where dA is the area element for where the force is applied. We make use of the symmetry of the problem and let dA be that of a ring of constant θ on the surface. The radius of the ring is r = R sin θ, where R is the radius of the sphere. If the angular width of the ring is dθ, in radians, then its width is R dθ and its area is dA = 2πR2 sin θ dθ. Thus the net horizontal component of the force of the air is given by

Fh=2πR2Δp\0π2sinθcosθdθ=πR2Δpsin2θ π/2=πR2Δp. 00

(b) We use 1 atm = 1.01 × 105 Pa to show that Δp = 0.90 atm = 9.09 × 104 Pa. The sphere radius is R = 0.30 m, so

Fh = π(0.30 m)2(9.09 × 104 Pa) = 2.6 × 104 N.

(c) One team of horses could be used if one half of the sphere is attached to a sturdy wall. The force of the wall on the sphere would balance the force of the horses.

8. We estimate the pressure difference (specifically due to hydrostatic effects) as follows: Δp=ρgh=(1.06×103 kg/m3)(9.8m/s2)(1.83m)=1.90×104 Pa.

9. Recalling that 1 atm = 1.01 × 105 Pa, Eq. 14-8 leads to ρgh=(1024kg/m3)(9.80m/s2)(10.9×103m)\0 1atm \0≈1.08×103 atm.

\01.01×105 Pa\0 \0\0

10. Note that 0.05 atm equals 5065 Pa. Application of Eq. 14-7 with the notation in this problem leads to

dmax = p =0.05atm=5065Pa. ρliquid g ρliquid g ρliquid g

Thus the difference of this quantity between fresh water (998 kg/m3) and Dead Sea water (1500 kg/m3) is

Δd = 5065 Pa \0 1 − 1 \0 = 5065 Pa \0 1 − 1 \0 = 0.17 m.

max g \0ρ ρ \0 9.8m/s2 \0998kg/m3 1500kg/m3 \0 \0fwsw\0\0 \0

11. The pressure p at the depth d of the hatch cover is p0 + ρgd, where ρ is the density of ocean water and p0 is atmospheric pressure. The downward force of the water on the hatch cover is (p0 + ρgd)A, where A is the area of the cover. If the air in the submarine is at atmospheric pressure then it exerts an upward force of p0A. The minimum force that must be applied by the crew to open the cover has magnitude

F = (p0 + ρgd)A – p0A = ρgdA = (1024 kg/m3)(9.8 m/s2)(100 m)(1.2 m)(0.60 m) =7.2×105 N.

12. With A = 0.000500 m2 and F = pA (with p given by Eq. 14-9), then we have ρghA = 9.80N. Thisgivesh≈2.0m,whichmeansd+h=2.80m.

13. In this case, Bernoulli’s equation reduces to Eq. 14-10. Thus,

pg =ρg(−h)=−(1800kg/m3)(9.8m/s2)(1.5m)=−2.6×104 Pa.

14. Using Eq. 14-7, we find the gauge pressure to be pgauge