Math  /  Data & Statistics

QuestionA source of sound emits 10.0 watts at the very centre of a spherical glass ball of radius 1.00 m . a) How much power passes through the whole sphere? b) What is the intensity of sound at the glass sphere? c) How much power passes through a 1.0 m21.0 \mathrm{~m}^{2} portion of the sphere?

Studdy Solution

STEP 1

What is this asking? We've got a sound source stuck inside a glass ball, and we want to figure out how much sound energy is passing through the whole ball, how intense the sound is at the surface, and how much energy goes through a small patch of the ball. Watch out! Don't mix up power and intensity!
Power is the *total* energy emitted, while intensity is how much power passes through a specific area.

STEP 2

1. Calculate the total power passing through the sphere.
2. Calculate the intensity at the sphere's surface.
3. Calculate the power through a portion of the sphere.

STEP 3

Alright, let's **start** with the **total power** passing through the sphere!
Since the sound source is right at the center of the glass ball, *all* the sound energy it emits has to pass through the sphere.
It's like blowing up a balloon – all the air you blow in has to stretch the balloon out!

STEP 4

So, the **total power** passing through the sphere is the same as the **power emitted by the source**, which is 10.0\bf{10.0} **watts**.
Easy peasy!

STEP 5

Now, let's talk about **intensity**.
Intensity is how much power is spread over a certain area.
Think of it like spreading butter on toast – if you have the same amount of butter, but a bigger piece of toast, the butter layer will be thinner, right?

STEP 6

The formula for **intensity** is I=PA I = \frac{P}{A} where II is the **intensity**, PP is the **power**, and AA is the **area**.

STEP 7

We know the **power** (P=10.0P = 10.0 watts) and we can find the **area** of the sphere because we know its **radius**!
The formula for the surface area of a sphere is A=4πr2 A = 4 \cdot \pi \cdot r^2 where rr is the **radius**.

STEP 8

The **radius** is 1.00\bf{1.00} **meter**, so the **area** is A=4π(1.00 m)2=4π1.00 m212.57 m2 A = 4 \cdot \pi \cdot (1.00 \text{ m})^2 = 4 \cdot \pi \cdot 1.00 \text{ m}^2 \approx 12.57 \text{ m}^2

STEP 9

Now, we can plug the **power** and **area** into the **intensity** formula: I=10.0 W12.57 m20.796 W/m2 I = \frac{10.0 \text{ W}}{12.57 \text{ m}^2} \approx 0.796 \text{ W/m}^2 So, the **intensity** at the sphere's surface is approximately 0.796\bf{0.796} W/m2\bf{\text{W/m}^2}.

STEP 10

Finally, let's figure out how much power goes through a 1.0\bf{1.0} m2\bf{\text{m}^2} patch of the sphere.
We know the **intensity** at the surface, and we know the **area** of the patch.
We can use the same **intensity** formula, but rearrange it to solve for **power**: P=IA P = I \cdot A

STEP 11

We found the **intensity** (I0.796 W/m2I \approx 0.796 \text{ W/m}^2) and the **area** of the patch is A=1.0 m2A = 1.0 \text{ m}^2, so P=0.796 W/m21.0 m2=0.796 W P = 0.796 \text{ W/m}^2 \cdot 1.0 \text{ m}^2 = 0.796 \text{ W}

STEP 12

a) The total power passing through the sphere is 10.0\bf{10.0} **watts**. b) The intensity of the sound at the glass sphere is approximately 0.796\bf{0.796} W/m2\bf{\text{W/m}^2}. c) The power passing through a 1.0 m21.0 \text{ m}^2 portion of the sphere is approximately 0.796\bf{0.796} **watts**.

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