Question

LEARN MORE REMARKS It's interesting that the motion of fluids can be treated with the same kinematics equations as individual objects. QUESTION By what factor would the range be changed if the flow rate were doubled? ***** * * ****** PRACTICE IT Use the worked example above to help you solve this problem. A water hose 2.20 cm in diameter is used by a gardener to fill a 33.0-liter bucket. (One liter = 1,000 cm?.) The gardener notices that it takes 1.00 min to fill the bucket. A nozzle with an opening of cross-sectional area 0.500 cm2 is then attached to the hose. The nozzle is held so that water is projected horizontally from a point 1.00 m above the ground. Over what horizontal distance can the water be projected? EXERCISE HINTS: GETTING STARTED I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. The nozzle is replaced with a Y-shaped fitting that splits the flow in half. Garden hoses are connected to each end of the Y, with each hose having a 0.350 cm? nozzle. (a) How fast does the water come out of one of the nozzles? m/s (b) How far would one of the nozzles squirt water if both were operated simultaneously and held horizontally 1.00 m off the ground? (Hint: Find the volume flow rate through each 0.350-cm2 nozzle, then follow the same steps as before.) Need Help? Read It Talk to a Tutor

Answer 1

Practice it :- Volume of bucket (v) = 33 e time (t) = 1 min xm ment = 60 sec. .. discharge = = 330 mm) Area of nozzle €A) = 0.5 cm² = 0.5810 m² . Velocity at outlet of nozzle = discharge a ( Area = 20,000 xos xoay 11 m/s s JV = 11 m/s Now, time require to reach water from nozzle outled to bucket .(t) to Sy= V t + tay t2 where, by displace in y-direction. 1 = 0 + 1 xgx t? Vy = initial velocity (vy=0 in yo direction t = 0.451 sec ay-acceleration ay = g) Now, in a direction. Sn = ut + 1/4 ax t2 sx = vxt [V=V sa = 11 x 0.45 5x = 4.961 m) Horizontal distance of water I Projected. ax=0 a nous zur yury at

Exercise: given, Area of nozzle (A) = or 18 an? 0. 35 cm² Q = 0.35x100 m2 Z A =0.250 cm ² Total discharge (@) = 11 m/s from one nozzle, discharge is 2 = 16 20000 ms y ll% = 1 2000g 2 40000' the Know that discharge = cross-section of area & Velocity. % = AX V = 0.35 x0x V 40000 V = '70 857. m/s Velocity from one Wozzle. lone nozzle (Top view) Sy = Vyd + Ź g t 2 (vy = 0) = Ź g +12, ar t = ( 2 y = 0.451 See be sx Now, Sn= Vxt + haut 2 sx - vrt' to (a =0) * Please like if you find & this solution helpful Thank you * 5x = 7.85780.451 * 15x = 3.5 44 m/ RUV