A tube has two area of cross- sections as shown in figure. The diameters of the tube are 8 mm and 2 mm. Find range of water falling on horizontal surface, if piston is moving with a constant velocity of 0.25 m/s, h = 1.25 m (g = 10 m/s 2 )

Question

Kevin Nash · Answer

Hello Student,

Please find the answer to your question

From law of continuity

A₁ v₁ = A₂ v₂

Given A₁ = π x (4 x 10^-3m)² , A₂ = π x (1 x 10^{-3 m})²

v₁ = 0.25 m/s

∴ v₂ = π x (4 x 10^-3)² x 0.25 / π x (1 x 10^-3)²

Also, h = 1 /2 gt² ⇒ t = √2h / g

Horizontal range x = v² √2h / g = 4 x √ 2 x 1.25 / 10 = 2 m

Thanks
Kevin Nash
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A tube has two area of cross- sections as shown in figure. The diameters of the tube are 8 mm and 2 mm. Find range of water falling on horizontal surface, if piston is moving with a constant velocity of 0.25 m/s, h = 1.25 m (g = 10 m/s 2 )

A tube has two area of cross- sections as shown in figure. The diameters of the tube are 8 mm and 2 mm. Find range of water falling on horizontal surface, if piston is moving with a constant velocity of 0.25 m/s, h = 1.25 m (g = 10 m/s²)

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A tube has two area of cross- sections as shown in figure. The diameters of the tube are 8 mm and 2 mm. Find range of water falling on horizontal surface, if piston is moving with a constant velocity of 0.25 m/s, h = 1.25 m (g = 10 m/s 2 )

A tube has two area of cross- sections as shown in figure. The diameters of the tube are 8 mm and 2 mm. Find range of water falling on horizontal surface, if piston is moving with a constant velocity of 0.25 m/s, h = 1.25 m (g = 10 m/s2)

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A tube has two area of cross- sections as shown in figure. The diameters of the tube are 8 mm and 2 mm. Find range of water falling on horizontal surface, if piston is moving with a constant velocity of 0.25 m/s, h = 1.25 m (g = 10 m/s²)