Question

guys.. pls help me out with this question​

Answer 1

Answer:

Exit velocity = 29.43 m/s, Mass Flow Rate = 2.31 kg/s

Explanation:

Given :

Exit diameter = 1 cm = 0.01 m

$A = \frac{\pi}{4}D_{e} ^2$

   = π/4 * 0.01²

   = 3.14/4 * 0.01²

   = 3.14/4 * (10⁻⁴)

   = 0.785 * 10⁻⁴

   = 7.85 * 10⁻⁵

Then :

Exit Velocity  will be :

$V_{e} = (\frac{-2 W v_{e}}{A})^{\frac{1}{3} }$

$\Rightarrow (\frac{2 * 1000 * 0.001001}{7.85 * 10^{-5} } )^{\frac{1}{3} }$

⇒ 29.43 m/s

Now,

Mass Flow rate will be :

$m = \frac{AV_{e}}{V_{e} }$

=> (7.85 * 10⁻⁵ * 29.43)/(0.001001)

=> 2.31 kg/s

Therefore :

Exit velocity = 29.43 m/s

Mass Flow Rate = 2.31 kg/s

Hope it helps!

Answer 2

Answer:

Exit velocity = 29.43 m/s, Mass Flow Rate = 2.31 kg/s

Explanation:

Given :

Exit diameter = 1 cm = 0.01 m

  = π/4 * 0.01²

  = 3.14/4 * 0.01²

  = 3.14/4 * (10⁻⁴)

  = 0.785 * 10⁻⁴

  = 7.85 * 10⁻⁵

Then :

Exit Velocity  will be :

⇒ 29.43 m/s

Now,

Mass Flow rate will be :

=> (7.85 * 10⁻⁵ * 29.43)/(0.001001)

=> 2.31 kg/s

Therefore :

Exit velocity = 29.43 m/s

Mass Flow Rate = 2.31 kg/s

__________________________________________________________

Hope it helps <3

Brainlist plz mate