Question

The next three questions concern water flowing in or from a hose. The cross section of the hose is circular. The circumference of a circle is 2TR and the area of a circle is R2, where Ris the radius the circle. 1 milliliter = 1 cm. Question 16 2 pts If the inner diameter of the hose is 1.24 cm, and the water flows in the hose with a speed of 0.82 m/s, what is the volumetric current Q of the water flowing in the hose (in milliliters per second)? Question 17 2 pts If the inner diameter of the hose is 1.2 cm, but you partially cover the opening of the hose leaving an open area of 0.16 cm²at the end of the hose so that you can spray your friends. The water flows out at 4.4 m/s. What is the volumetric current in milliliters per second now?

Answer 1

16) P = AV - 3.14 4 x{1•24) x 82 = 99.0 cm? 4 (Ad JV a = 99.0 millimetes per second 2 17) A = 0.16 cm v=4.4 m/s = 440cm/s Q = Av = 0.16 X 440 = 70.4 millimeter/s A, V 18) Equation of continuity. Ą ķ (ad') v = (0-3)(3-1) → V = 0.3 X 3-584 0.929 m/s 3-14 X (1.2)²

Answer 2

SOLUTION 

16.

Q = A * v = pi/4 (1.24)^2 * 0.82*100. cm^3/s (milliliteres per sec)

=> Q = 99.03 milliliters/sec (ANSWER).

17.

Volumetric current remains unchanged.

So.

Volumetric current after partial blocking or before

= 0.16 * 4.4 * 100  

= 70.4 millilitres/sec (ANSWER)

18.

Let speed of water flowing into hose be v0 m/s.

As flow rate remains unchanged at entry and exit of hose.

Q (entry) = Q(exit)

=> pi/4 (1.2)^2 * v0 * 100   =  (0.3) * 3.5 * 100 

=> v0 = 0.3 * 3.5  *4/(pi*(1.2)^2) = 0.9284 m/s (ANSWER)