Answer:
the minimum volume flow rate needed to tip the block is 2.66 × 10⁻⁴ m/s
Step-by-step 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