Water is flowing through a rectangular channel that is 12 m wide with a speed of 0.75 m/s. The water then flows into four identical rectangular channels that have a width of 3.0 m. The depth of the water does not change as it flows into the four channels. What is the speed of the water in one of the smaller channels?

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Bot · Answer

We can use the principle of conservation of flow rate in this problem. The flow rate in the main channel is equal to the sum of the flow rates in the four smaller channels. Flow rate is given by the product of the cross-sectional area of the channel and the speed of the water flow.

Let's denote the width of the main channel as W1 (12 m), the width of each smaller channel as W2 (3 m), the speed of water in the main channel as V1 (0.75 m/s), and the speed of water in the smaller channel as V2.

The depth of the water remains the same, so we'll just use a generic term, D, to represent the depth of the water.

We can calculate the flow rate in the main channel:
Flow_rate_main_channel = Area_main_channel × V1
Flow_rate_main_channel = (W1 × D) × V1
Flow_rate_main_channel = (12 × D) × 0.75

Now let's calculate the flow rate in one of the smaller channels:
Flow_rate_one_small_channel = Area_one_small_channel × V2
Flow_rate_one_small_channel = (W2 × D) × V2
Flow_rate_one_small_channel = (3 × D) × V2

There are 4 smaller channels, and the flow rate in the main channel is equal to the total flow rate in the smaller channels. So, we can express this as:

Flow_rate_main_channel = 4 × Flow_rate_one_small_channel

Now we can plug in the expressions for the flow rates that we found earlier:

(12 × D) × 0.75 = 4 × (3 × D) × V2

We can see that D is present on both sides of the equation, so we can divide both sides by it:

(12 × 0.75) = 4 × (3) × V2

Next, we can solve for V2:

9 = 12 × V2
V2 = 9 / 12
V2 = 0.75 m/s

So, the speed of the water in one of the smaller channels will be 3 m/s.