A 10 mm diameter jet of water is deflected by a homogeneous rectangular block (15 mm by 200 mm by 100 mm) that weighs 6 N. Determine the minimum volume flowrate needed to tip the block.

the minimum volume flow rate needed to tip the block is 2.66 × 10⁻⁴ m/s

Explanation:

Given that;

diameter of the jet d = 10 mm

weight W = 6 N

Now we say

Fₓ Lfₓ - Wlw= 0

horizontal force

Fₓ = W (lw/lfₓ)

Fₓ = 6 ( 0.015/2)

Fₓ = 0.9 N

X-component of momentum

v₁p(-v₁)A₁ = - Fₓ

pA₁v₁² = Fₓ

v₁² = Fₓ / pA₁

v₁ = √( Fₓ / pA₁ )

WE SUBSTITUTE

v₁ = √ ( 0.9 / ((999)(π/4)(0.01))²

v₁ = 3.39 m/s

Now Discharge Q = A₁v₁

Q = π/4 (0.01)² (3.39)

Q = 2.66 × 10⁻⁴ m/s

therefore the minimum volume flow rate needed to tip the block is 2.66 × 10⁻⁴ m/s

answer: eco and eco system

explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

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