I need 16 ..ooo AT&T; 2:25 PM quest.cns.utexas.edu 3 of 3 sconce (las849)-HW 7-s

Question

I need 16 ..ooo AT&T; 2:25 PM quest.cns.utexas.edu 3 of 3 sconce (las849)-HW 7-spurlock (44103) is the ce n a porthole of diameter 23.6 em? Answer in units of N 014 10.0 points .2 ata 5.71 m 185 ms 0.13 m The small piston of a hydranlic lift has a cross-sectional area of 7.4 cm and the large piston has an area of 62 em2, as in the figure below 0.05 m Applying Bernoulli's principle, what is the pressure Pi at the entrance end of the pipe? Assume the viscosity of the uid is negligible and the Buid is incompressible. The acceleration of gravity is 9.8 m/ and Pan" 1.013 x 10', Pa. 62 cm2 Answer in units of Pa 017 10.0 points The hypodermic syringe contains a medicine What force F must be applied to thel with the same density of water. The barrel has a cross-sectional area of piston to maintain the load of 78 kN at a constant elevation? of the 3.7531 × 10-5 · The cross-sectional area of the nodle is 2.56679 × 10 8 1, In the absenoe of a force on the plunger, the pressure everywhere is 1.0 atm. A force of magnitude Answer in units of N 015 10.0 points A ferry boat is 4.3 m wide and 7.8 m long. 2.65935 N is exerted on the plunger, making When a truck pulls outo it, te bost sinks medicine squirt from t 6.52 cm in the water medicine squirt from the needle. What is the weight of the truck? The acceleration of gravity is 9.81 m/s Answer in units of N 016 10.0 points A liquid of density 1202 kg/fows with speed 1 ,85 m/s into a pipe of diameter 0.13 m The diameter of the pipe decreases to 0.05 m Find the medicine's flow speed through the to 0.05 needle. Assume t dle remains at that the pressure pressure in the e- at its exit end. The exit end of the pipe is remains at atmospheric pressure, that the 5.71 m lower than the entrance of the pipe, syringe is horizontal, and that the speed of and the pressure at the exit of the pipe is the emerging fluid is the same as the speed of 1.2 atm the fluid in the needle, Answer in units of m/s

Explanation / Answer

given that liquid has density rh0 = 1202 kg /m^3

speed is v1 = 1.85 m/s

diameter d1 = 0.13 m

area of cross section is A1 = pi*(d1/2)^2 = 3.142*(0.13/2)^2 = 0.01327 m^2

new diameter is d2 = 0.05 m

new area of cross section is A2 = pi*(d2/2)^2 = 3.142*(0.05/2)^2 = 0.00196 m^2

h1-h2 = 5.71 m

p2 = 1.2 atm = 1.2*1.013*10^5 = 1.2156*10^5 Pa

P1 = ?

use equation of continuty A1*v1 = A2*v2

v2 = (A1*v1/A2) = ( 0.01327*1.85)/( 0.00196 ) = 12.52 m/s

According to bernoullies theorem

P1+(0.5*rho*v1^2)+ (rho*g*h1) = P2+ (0.5*rho*v2^2) +(rho*g*h2)

P1 = P2+(0.5*rho*(v2^2-v1^2)) + (rho*g*(h2-h1))

P1 = 1.2156*10^5 + (0.5*1202*( 12.52^2 - 1.85^2) -(1202*9.81*5.71)

P1 = 121560 + 92150.1 -(67330.15) = 146379.95 Pa = 1.46*10^5 Pa

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.