To solve the problem, we use the principles of fluid dynamics, specifically Bernoulli’s equation and the continuity equation. Here’s a detailed breakdown:

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Step 1: Bernoulli’s Equation

Bernoulli’s equation states: $P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2$ Since the liquid exits horizontally and there’s negligible height difference, the terms involving $g h$ cancel out. Additionally, the liquid starts at rest inside the syringe, so $v_1 \approx 0$. The equation simplifies to: $P_1 = P_2 + \frac{1}{2} \rho v_2^2$ Rearranging to solve for $v_2$, the speed of the liquid exiting the needle: $v_2 = \sqrt{\frac{2 (P_1 - P_2)}{\rho}}$

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Step 2: Calculate Pressure Difference

The force $F$ applied to the plunger creates an additional pressure in the syringe: $P_\text{applied} = \frac{F}{A}$ where:

• $F = 2.30 \, \text{N}$
• $A = 2.63 \times 10^{-5} \, \text{m}^2$

Substituting: $P_\text{applied} = \frac{2.30}{2.63 \times 10^{-5}} = 8.75 \times 10^4 \, \text{Pa}$

The total pressure inside the syringe becomes: $P_1 = P_\text{atm} + P_\text{applied}$ $P_1 = 1.00 \, \text{atm} + 8.75 \times 10^4 \, \text{Pa}$ Converting $1.00 \, \text{atm}$ to Pascals ($1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa}$): $P_1 = 1.013 \times 10^5 + 8.75 \times 10^4 = 1.888 \times 10^5 \, \text{Pa}$

The pressure at the exit ($P_2$) is atmospheric pressure: $P_2 = 1.013 \times 10^5 \, \text{Pa}$

The pressure difference is: $\Delta P = P_1 - P_2 = 1.888 \times 10^5 - 1.013 \times 10^5 = 8.75 \times 10^4 \, \text{Pa}$

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Step 3: Solve for Speed $v_2$

The density of the solution is the same as water ($\rho = 1000 \, \text{kg/m}^3$). Substituting into the velocity equation: $v_2 = \sqrt{\frac{2 \Delta P}{\rho}}$ $v_2 = \sqrt{\frac{2 (8.75 \times 10^4)}{1000}}$ $v_2 = \sqrt{175} \approx 13.23 \, \text{m/s}$

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Final Answer:

The speed of the solution as it leaves the needle is approximately: $\boxed{13.2 \, \text{m/s}}$

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Would you like further clarification or more examples?

Related Questions:

1. How does the viscosity of the liquid affect the flow rate in this scenario?
2. What assumptions were made in this calculation, and how could they impact the result?
3. How would the speed change if the cross-sectional area of the plunger was doubled?
4. What is the relationship between the applied force and the exit speed of the liquid?
5. How does Bernoulli’s equation apply to non-horizontal fluid movement?

Tip:

Always ensure consistent units throughout calculations to avoid errors in pressure or velocity evaluations!