What kind of relationship exists between sound and the temperature of material

We can reasonably model a 90-W incandescent lightbulb as a sphere 5.7 cm in diameter. Typically, only about 5% of the energy goe

o-na [289]

Answer:

See answer

Explanation:

Given quantities:

$\eta = 0.05\\ W=90[W]\\r=0.0285[m]$

where $\eta$ is the efficiency of the lightbulb (visible light is 5% of the total power), $W$ is the total power of the lightbulb, r is the radius of the lightbulb in meters.

Intensity is power divided by area:

$I =\frac{P}{A}$

a) Now the effective power is $\eta*W$ , therefore:

$I =\frac{\eta*W}{\pi r^2}=\frac{0.05*90}{4\pi (0.0285)^2}=440.87[W/m^2]$

b) Now the intensity is the average poynting vector is related to the magnitudes of the maximum electric field and magnetic field amplitudes, following:

$S_{average}= \frac{EB}{2\mu_{0}}[W/m]$

now $E$ and $B$ are related:

$E=cB\\ B=\frac{E}{c}$ and $c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}$

replace in $S_{average}$

$S_{average}=I= \frac{c \epsilon_{0}E^2}{2}[W/m]$

we replace the values and we get:

$E= \sqrt{\frac{2I}{\epsilon_{0}c}}$

$E = \sqrt{\frac{2(440.8)}{8.85*10^{-12}3*10^8}}=576.24[V/m]$

therefore

$B=\frac{E}{c}=\frac{576.24}{3*10^{8}}=1.92*10^{-6}[T]$

8 0

3 years ago

What do you do , when you get anger?

Vera_Pavlovna [14]

I try to think of things that calms me down. If that doesn't work, I try to think of things that makes me happy. As I once read in a book, you can never get over your anger but you can calm yourself down by forgetting about it.<span />

6 0

3 years ago

Read 2 more answers

Are materials that dissolve in water hydrophobic or hydrophilic

stira [4]

Answer:

hydrophilic

Explanation:

hydrophobic means it hates water so a hydrophobic material would separate from the water and just sit there (an example of this is oil)

4 0

3 years ago

Read 2 more answers

If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm,

qaws [65]

Answer:

the natural length of the spring is 9 cm

Explanation:

let the natural length of the spring = L

For each of the work done, we set up an integral equation;

$5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2} \ \ \ -----(1)$

The second equation of work done is set up as follows;

$9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)$

solve equation (1) and equation (2) together;

$\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12} (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\$

$l = 9 \ cm$

Therefore, the natural length of the spring is 9 cm

4 0

3 years ago

One molecule of glucose makes 30 molecules of atp. how many molecules of glucose are needed to make 300 molecules of atp in aero

leonid [27]

10 molecules are required

8 0

4 years ago

Read 2 more answers