9. A hole of radius 1.0 mm occurs in the bottom of a water storage tank that hol

Question

9. A hole of radius 1.0 mm occurs in the bottom of a water storage tank that holds water to a depth of 15 m. At what rate will water flow out of the hole? (Density of water is 1 kg/L, 1 m3 = 1000 L)

A) 5.4 mL/s

B) 54 mL/s (i got this)

C) 540 mL/s

D) 0.54 mL/s

10. When you blow some air above a paper strip, the paper rises. This is because

A) the air above the paper moves slower and the pressure is higher.

B) the air above the paper moves faster and the pressure is higher.

C) the air above the paper moves faster and the pressure is lower. ( i got this)

D) the air above the paper moves slower and the pressure is lower.

11. The flow rate of blood through the average human aorta, of radius 1.0 cm, is about 90 cm3/s. What is the speed of the blood flow through the aorta?

A) 29 cm/s ( i got this)

B) 37 cm/s

C) 32 cm/s

D) 14 cm/s

12. An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. The speed of the fluid flow through the 3.0-inch section will be what factor times that through the 1.0-inch section?

A) 1/3

B) 9.0

C) 6.0

D) 1/9 (i got this)

Explanation / Answer

linear speed of leakage through a narrow hole is particular case of Bernoulli equation v=?(2gh);
? rate of water volume loss is R=v*A, where A=pi*r^2 is area of the hole; thus
R=?(2gh)*pi*r^2 = ?(2*9.8*15) *pi*1^2 =54 mL/s

2.ans C

.the air above the paper moves faster and the pressure is lower

.It's the Bernoulli effect.

3.ans A

Remember, Flow = Velocity * Cross area.
V = F / A
They only gave you radius, not cross sectional area, but that's just ?(r^2) or 3.14* (1^2)
A = 3.14 cm2
F = 90 cm3/s

V = 90 / 3.14
V = 28.66 cm/s appprox to 29

4.ans D

This doesn't have anything to do with buoyancy.

Since the area of the large section is 9 times that of the small section, the velocity must be 1/9th for each to carry the same amount of fluid per unit time.

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