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Math  /  Algebra

QuestionWater enters a shower head through a cylindrical pipe with an inner radius of a=1.27 cma=1.27 \mathrm{~cm} at a speed of vin=0.20 m/sv_{i n}=0.20 \mathrm{~m} / \mathrm{s}. The water exits the shower head through 30 identical circular holes each with a radius of b=0.50b=0.50 millimeters. What is the speed Vout V_{\text {out }} of the water exiting the holes in the shower head? Assume that the incoming water is divided equally between the 30 holes. You may neglect all friction and ignore any work done on the water by gravity.

Studdy Solution

STEP 1

What is this asking? How fast does water shoot out of a shower head with 30 tiny holes, if it enters slowly through a larger pipe? Watch out! Don't mix up the units, centimeters and millimeters!
Also, remember the water splits between *all* the holes.

STEP 2

1. Calculate the pipe's area.
2. Calculate a single hole's area.
3. Calculate the total area of all holes.
4. Find the speed of the water exiting the holes.

STEP 3

Alright, first things first!
We need the area of that pipe.
It's a circle, right?
And the area of a circle is π⋅radius2\pi \cdot \text{radius}^2.
The problem says the pipe's radius is a=1.27a = 1.27 **cm**.
Let's convert that to **meters**: 1.27 cm⋅1 m100 cm=0.01271.27 \text{ cm} \cdot \frac{1 \text{ m}}{100 \text{ cm}} = 0.0127 **m**.

STEP 4

Now, plug that radius into the area formula: Ain=π⋅(0.0127 m)2=π⋅0.00016129 m2≈0.0005067 m2 A_{in} = \pi \cdot (0.0127 \text{ m})^2 = \pi \cdot 0.00016129 \text{ m}^2 \approx 0.0005067 \text{ m}^2 So, the **pipe's area** is approximately 0.0005067 m20.0005067 \text{ m}^2.

STEP 5

Now, let's find the area of one of those tiny holes.
The radius of each hole is b=0.50b = 0.50 **mm**.
We need to convert that to **meters**: 0.50 mm⋅1 m1000 mm=0.00050.50 \text{ mm} \cdot \frac{1 \text{ m}}{1000 \text{ mm}} = 0.0005 **m**.

STEP 6

Using the same circle area formula, the area of one tiny hole is: Ahole=π⋅(0.0005 m)2=π⋅0.00000025 m2≈0.0000007854 m2 A_{hole} = \pi \cdot (0.0005 \text{ m})^2 = \pi \cdot 0.00000025 \text{ m}^2 \approx 0.0000007854 \text{ m}^2

STEP 7

We've got 30 of these little holes, and the water is splitting between them.
So, the total area of all the holes is: Aout=30⋅Ahole=30⋅0.0000007854 m2≈0.00002356 m2 A_{out} = 30 \cdot A_{hole} = 30 \cdot 0.0000007854 \text{ m}^2 \approx 0.00002356 \text{ m}^2

STEP 8

Here's the key idea: the amount of water flowing in has to equal the amount flowing out!
We can write that as: Ain⋅vin=Aout⋅vout A_{in} \cdot v_{in} = A_{out} \cdot v_{out} Where vinv_{in} is the speed of the water coming *in*, and voutv_{out} is the speed going *out*.
This is called the **continuity equation**, and it's super important in fluid dynamics!

STEP 9

We know AinA_{in}, vinv_{in}, and AoutA_{out}, so we can solve for voutv_{out}: vout=Ain⋅vinAout v_{out} = \frac{A_{in} \cdot v_{in}}{A_{out}} Plugging in our values: vout=0.0005067 m2⋅0.20 m/s0.00002356 m2≈4.3 m/s v_{out} = \frac{0.0005067 \text{ m}^2 \cdot 0.20 \text{ m/s}}{0.00002356 \text{ m}^2} \approx 4.3 \text{ m/s}

STEP 10

The water blasts out of the shower head at a speed of approximately 4.34.3 **m/s**!

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