Question

A small hole of area of cross-se
m Taking g = 10 m/s2, the rate
of area of cross-section 2 mm- is present near the bottom of a fully filled open tank of height 2
= 10 m/s2, the rate of flow of water through the open hole would be nearly​

Answer 1

Answer:

hAdg= by substituting in this formula u get answer

Answer 2

Question: A small hole of area of cross-section 2 $mm^{2}$ is  present near the bottom of a fully filled open  tank of height 2 m. Taking g = 10 $m/s^{2}$, the rate  of flow of water through the open hole would  be nearly:

Solution: $12.6 \times 10^{-6} m^{2}/s$

Step by step explanation:

$Q=au=a\sqrt{2gh}\\ =2 \times 10^{-6} m^{2} \times\sqrt{2 \times 10 \times 2} m/s\\= 2 \times 2 \times 3.14 \times 10^{-6}m^{3}/s$

Therefore the rate of flow through the open hole would be,

$:\implies \boxed{12.6 \times 10^{-6} m^{3}/s}$