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3 years ago

What is the frequency of the most intense radiation emitted by your body? assume a skin temperature of 95 ∘f?

1 answer:

7 0

Answer:
Wien's law:
Î»_peak = b/T
Wien's constant: b = 2.8977685(51)Ă—10â’3 mâ€˘K
T = (5/9)[96 â€“ 32) + 273 = 35.55 + 273 = 308.55 deg. K
Î»_peak = 2.8977685(51)Ă—10â’3 /308.55 = 9.39x10^-6 = 9.39 um

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An unbalanced force of 20N acts on a 4.0kg mass what is it's acceleration

tankabanditka [31]

Hi there!

According to Newton's second law:

∑F = m · a, where:

∑F = net force (N = kgm/s²)

m = mass (kg)

a = acceleration (m/s²)

Rearrange to solve for acceleration:

F/m = a

20N / 4.0kg = 5 m/s²

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3 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

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3 years ago

What cause earthquakes?

KengaRu [80]

Question; what causes earthquakes?
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4 years ago

Read 2 more answers

What’s an example of contact friction

kotegsom [21]

Answer:

a nightstand on a lamp table

Explanation:

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3 years ago

Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p

PSYCHO15rus [73]

Answer:

Energy density will be 14.73 $J/m^3$

Explanation:

We have given capacitance $C=225\mu F=225\times 10^{-6}F$

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm $0.2\times 10^{-3}m$

We know that there is relation between electric field and potential

$E=\frac{V}{d}$ , here E is electric field, V is potential and d is separation between the plates

So $E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C$

Energy density is given by $E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3$

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3 years ago