Answer:
New speed of water is 2.8 m/s.
Step-by-step explanation:
It is given that,
Speed of water, v₁ = 2.3 m/s
Diameter of pipe, d₁ = 3.2 cm
Radius of pipe, r₁ = 1.6 cm = 0.016 m
Area of pipe,
![A_1=\pi(0.016)^2=0.000804\ m^2](https://img.qammunity.org/2020/formulas/physics/college/xgi7mqppp0627qq6okmxz0m0duu1udsu98.png)
If the pipe narrows its diameter, d₂ = 2.9 cm
Radius, r₂ = 0.0145 m
Area of pipe,
![A_2=\pi(0.0145)^2=0.00066\ m^2](https://img.qammunity.org/2020/formulas/physics/college/44uz01p544x6dkmm5xcah9ah6k78psz8ip.png)
We need to find the new speed of the water. It can be calculated using equation of continuity as :
![v_1A_1=v_2A_2](https://img.qammunity.org/2020/formulas/physics/college/w7pv8sr0b9qrb9ivz4csy4lu94nqmhyb1k.png)
![v_2=(v_1A_1)/(A_2)](https://img.qammunity.org/2020/formulas/physics/college/6wfbtxpm94pytidxwdtzbei0fl5kmqexgt.png)
![v_2=(2.3\ m* 0.000804\ m^2)/(0.00066\ m^2)](https://img.qammunity.org/2020/formulas/physics/college/mudg8rkj7uw79fjjgmvv9u1ywml9s7yn3l.png)
![v_2=2.8\ m/s](https://img.qammunity.org/2020/formulas/physics/college/v6pxgtsmjlmh88ch6qfsh6xcui1qrj3va9.png)
So, the new speed of the water is 2.8 m/s. Hence, this is the required solution.