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777dan777 [17]
2 years ago
10

As a rough approximation, the human body may be considered to be a cylinder of length L=2.0m and circumference C=0.8m. (To simpl

ify things, ignore the circular top and bottom of the cylinder, and just consider the cylindrical sides.) If the emissivity of skin is taken to be e=0.6, and the surface temperature is taken to be T=30∘C, how much thermal power P does the human body radiate?
Physics
1 answer:
Brilliant_brown [7]2 years ago
4 0

Answer:

Thermal Power = 460W

Explanation:

From Stephan-Boltzmann Law Formula;

P = єσT⁴A

Where,

P = Radiation energy

σ = Stefan-Boltzmann Constant

T = absolute temperature in Kelvin

є = Emissivity of the material.

A=Area of the emitting body

Now, σ = 5.67 x 10^(-8)

є = 0.6

Temperature = 30°C and coverting to kelvin = 30 + 273 = 303K

Area ; since we are to consider the sides of the human body as 2m and 0.8m,thus area = 2 x 0.8 = 1.6

Thus thermal power = 0.6 x 5.67 x 10^(-8) x303⁴ x 1.6 = 458. 8W

Normally, we approximate to the nearest 10W. Thus, thermal power is approximately 460W

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lord [1]

Answer:

To both observers, the land opposite them is moving to the right.

Explanation:

I have this class too. and there is also a quizlet with all the answers to the rest of the other questions. Trust me its right .

https://quizlet.com/261219090/oce-1001-chapter-2-flash-cards/

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2 years ago
A thin stream of water flows smoothly from a faucet and falls straight down. at one point the water is flowing at a speed of v1
kati45 [8]
<span>The formulas are, v1d1² = v2d2² ........ (1) h = (v2²-v1²)/2g ...... (2) Given that, v1 = 1.71 m/s we assume that the stream has decreased by a factor d2 =0.805d1 then, v1d1² = v2 (0.805d1)² cancelled both side d1² then we get, v1 = v2 (0.805)² v1 = v2 (0.648025) Sub v1 = 1.71, 1.71 = v2 (0.648025) v2 = 1.71/0.648025 v2 = 2.638787083831642 v2 = 2.64 m/s The vertical distance formula, h = (v2²-v1²)/2g We know that value of gravity constant is 9.8 m/s² h = {(2.64)² - (1.71)²)/2(9.8) h = {(6.9696) - (2.9241)}/19.6 h = (4.0455)/19.6 h = 0.2064030612244898 h = 0.21 cm Therefore, the vertical distance h = 0.21 cm.</span>
8 0
2 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.80 μC/m and lies along the x-axis. A second long uniform li
Elodia [21]

Answer:

a) E=228391.8 N/C

b) E=-59345.91N/C

Explanation:

You can use Gauss law to find the net electric field produced by both line of charges.

\int \vec{E_1}\cdot d\vec{r}=\frac{\lambda_1}{\epsilon_o}\\\\E_1(2\pi r)=\frac{\lambda_1}{\epsilon_o}\\\\E_1=\frac{\lambda_1}{2\pi \epsilon_o r_1}\\\\\int \vec{E_2}\cdot d\vec{r}=\frac{\lambda_2}{\epsilon_o}\\\\E_2=\frac{\lambda_2}{2\pi \epsilon_o r_2}

Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.

The net electric field at point r will be:

E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})

a) for y=0.200m, r1=0.200m and r2=0.200m:

E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C

b) for y=0.600m, r1=0.600m, r2=0.200m:

E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C

5 0
2 years ago
Which of the following states that all matter tends to "warp" space in its vicinity and that objects react to this warping by ch
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There are no choices on the list you provided that make such a statement,
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That statement is one way to describe the approach to 'forces of gravity'
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Answer:

Explanation:

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So, the bullet is in contact with the plastic for the time,  \bigtriangleupt = \frac{(v - u)}{a} =\frac{(90 - 300)}{(-2047500)} = 1.03 \times 10^{-4} s

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8 0
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