The speed and pressure at another end will be 9 m/sec and -31500 pascals.
The difference between absolute pressure and atmospheric pressure is known as gauge pressure. Relative pressure is another name for gauge pressure.
Given data in problem is;
For horizontal pipe, Z₁=Z₂
Pressure at end 1, P₁ = 4500Pa
Speed at end 1, V₁ = 3 m/sec
Speed at end 2, V₂ = ? m/sec
Area of cross-section at end 1,A₁ =A
Area of cross-section at end 1,A₂ = A/3
From the continuity equation;
A₁V₁=A₂V₂
A× 3 = (A/3)×V₂
V₂ = 9 m/sec
From Bernoulli's equation;
[tex]\rm \frac{P_1}{\rho g} + \frac{v_1^2}{2g} +Z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} +Z_2 \\\\ \frac{P_1-P_2}{\rho g } = \frac{v_2^2-v_1^2}{2g} \\\\ \frac{P_1-P_2}{\rho} = \frac{v_2^2-v_1^2}{2} \\\\ \frac{4500-P_2}{1000} = \frac{9^2-3^2}{2} \\\\ 4500 -P_2 = 36000 \\\\ P_2 = - \ 31,500 \ pa[/tex]
Hence the speed and pressure at another end will be 9 m/sec and -31500 pascals.
To learn more about the gauge pressure, refer to the link;
https://brainly.com/question/14012416
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