Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.573 × 10–3 m³/s and the diameter of the nozzle you hold is 5.09 × 10-3 m. At what speed v does the water exit the nozzle?

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Chapter1: Units, Trigonometry. And Vectors
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**Problem Statement:**

Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of \(0.573 \times 10^{-3} \, \text{m}^3/\text{s}\) and the diameter of the nozzle you hold is \(5.09 \times 10^{-3} \, \text{m}\). At what speed \(v\) does the water exit the nozzle?

\[ v = \, \_\_\_\_\_ \, \text{m/s} \]

**Analysis:**

To solve this problem, we need to consider the relationship between the flow rate, nozzle diameter, and the speed of the water exiting the hose. The flow rate \(Q\) is given by the formula:

\[ Q = A \times v \]

where:
- \(Q\) is the flow rate,
- \(A\) is the cross-sectional area of the nozzle,
- \(v\) is the velocity of the water exiting the nozzle.

The cross-sectional area \(A\) of the nozzle can be calculated using the formula for the area of a circle:

\[ A = \pi \left(\frac{d}{2}\right)^2 \]

where \(d\) is the diameter of the nozzle. Here, \(d = 5.09 \times 10^{-3} \, \text{m}\).

Substitute the known values into the formulas to find \(v\), the speed of the water exiting the nozzle.
Transcribed Image Text:**Problem Statement:** Suppose you spray your sister with water from a garden hose. The water is supplied to the hose at a rate of \(0.573 \times 10^{-3} \, \text{m}^3/\text{s}\) and the diameter of the nozzle you hold is \(5.09 \times 10^{-3} \, \text{m}\). At what speed \(v\) does the water exit the nozzle? \[ v = \, \_\_\_\_\_ \, \text{m/s} \] **Analysis:** To solve this problem, we need to consider the relationship between the flow rate, nozzle diameter, and the speed of the water exiting the hose. The flow rate \(Q\) is given by the formula: \[ Q = A \times v \] where: - \(Q\) is the flow rate, - \(A\) is the cross-sectional area of the nozzle, - \(v\) is the velocity of the water exiting the nozzle. The cross-sectional area \(A\) of the nozzle can be calculated using the formula for the area of a circle: \[ A = \pi \left(\frac{d}{2}\right)^2 \] where \(d\) is the diameter of the nozzle. Here, \(d = 5.09 \times 10^{-3} \, \text{m}\). Substitute the known values into the formulas to find \(v\), the speed of the water exiting the nozzle.
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