Concept:For a horizontal pipe, the potential energy term (\( \rho g h \)) in Bernoulli's equation is zero, leaving only pressure and kinetic energy.
Formula:$$ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \implies \Delta P = P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2) $$
Solution:- We have \( \rho = 1000 \text{ kg/m}^3 \), \( v_1 = 1.0 \text{ m/s} \), \( v_2 = 3.0 \text{ m/s} \).
- Calculate the difference in the squares of the speeds: \( (3.0)^2 - (1.0)^2 = 9 - 1 = 8 \text{ m}^2/\text{s}^2 \).
- Multiply by half the density: \( \Delta P = \frac{1}{2} \times 1000 \times 8 \).
- \( \Delta P = 500 \times 8 = 4000 \text{ N/m}^2 \).
Why other options are incorrect:Calculating \( (v_2 - v_1)^2 \) instead of \( (v_2^2 - v_1^2) \) gives \( (3-1)^2 = 4 \), leading to 2000. Forgetting the \( 1/2 \) multiplier leads to 8000.
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