2, A garden hose of inner diameter 2.0 cm carries water at 2.0m/s The nozzle at the end has diameter 0.4 cm. how fast does the water move through the nozzle ? Ans: 50m/s

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2, A garden hose of inner diameter 2.0 cm carries water at 2.0m/s The nozzle at the end has diameter 0.4 cm. how fast does the water move through the nozzle ? Ans: 50m/s

Answer 1

### $50 \, \text{m/s}$

Answer 2

## Step 1: Identify the Problem ### We are given a problem dealing with the flow of water through a garden hose and a nozzle. The principle that applies here is the equation of continuity in fluid dynamics. This equation states that the product of the cross-sectional area ($A$) of a pipe and the velocity ($v$) of the fluid flow remains constant. ## Step 2: Convert Diameters to Radii ### The inner diameter of the garden hose is $2.0 \, \text{cm}$, so the radius $r_1$ is $1.0 \, \text{cm}$ or $0.01 \, \text{m}$. The diameter of the nozzle is $0.4 \, \text{cm}$, so the radius $r_2$ is $0.2 \, \text{cm}$ or $0.002 \, \text{m}$. ## Step 3: Calculate Cross-Sectional Areas ### The cross-sectional area of a tube or pipe is calculated as $A = \pi r^2$. Therefore, the areas are: \[A_1 = \pi r_1^2 = \pi (0.01)^2\] \[A_2 = \pi r_2^2 = \pi (0.002)^2\] ## Step 4: Apply the Equation of Continuity ### The equation of continuity is $A_1 v_1 = A_2 v_2$. We need to solve for $v_2$: \[v_2 = \frac{A_1 v_1}{A_2}\] Substituting the areas: \[v_2 = \frac{\pi (0.01)^2 \cdot 2.0}{\pi (0.002)^2}\] ## Step 5: Simplify and Solve ### Simplifying the expression: \[v_2 = \frac{(0.01)^2 \cdot 2.0}{(0.002)^2}\] \[v_2 = \frac{0.0001 \cdot 2.0}{0.000004}\] \[v_2 = \frac{0.0002}{0.000004}\] \[v_2 = 50 \, \text{m/s}\]

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2, A garden hose of inner diameter 2.0 cm carries water at 2.0m/s The nozzle at the end has diameter 0.4 cm. how fast does the water move through the nozzle ? Ans: 50m/s

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