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zavuch27 [327]
1 year ago
15

What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22

.0ºC . The person has a normal skin temperature of 33.0ºC and a surface area of 1.50 m2. The emissivity of skin is 0.97 in the infrared, where the radiation takes place.
Physics
1 answer:
SIZIF [17.4K]1 year ago
3 0

Answer:

5.45\times 10^{-4} W

Explanation:

T_{r} = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

T_{s} = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

A = Surface area = 1.50 m²

\epsilon = emissivity = 0.97

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})

R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})

R = 5.45\times 10^{-4} W

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The index of refraction for silicate flint glass is 1.66 for violet light that has a wavelength in air equal to 400 nm and 1.61
nikitadnepr [17]

Answer:

(a) Angle of incidence for violet is more than the angle of incidence for red

(b) 2.4°

Explanation:

refractive index for violet , v = 1.66

refractive index for red, nR = 1.61

wavelength for violet, λv = 400 nm

wavelength for red, λR = 700 nm

Angle of refraction, r = 30°

(a) Let iv be the angle of incidence for violet.

Use Snell,s law

nv = Sin iv / Sin r

1.66 = Sin iv / Sin 30

Sin iv = 0.83

iv = 56°

Use Snell's law for red

nR  = Sin iR / Sin r  

where, iR be the angle of incidence for red

1.61 = Sin iR / Sin 30

Sin iR = 0.805

iR = 53.6°

So, the angle of incidence for violet is more than red.

(b) iv - iR = 56° - 53.6° = 2.4°

4 0
1 year ago
Randy wants to know whether a soil's porosity affects how easily seedlings grow in it.
telo118 [61]
Since the main objective of this experiment is to determine the effect of porosity on seedling growth that should be the only independent variable. In short, that is the only variable that should be different to ensure fair testing. 

The answer should be B:

he plants seedlings in soils with different levels of porosity and equal levels of permeability. 

Permeability is not what needs to be tested. If it changes, you may not be able to determine whether it was the porosity or permeability that cause changes.  
4 0
2 years ago
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A metal sphere of radius 2.0 cm carries an excess charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sph
Nonamiya [84]

Answer:

The electric field is  E = 7.5 *10^{6} \ N/C

Explanation:

From the question we are told that

    The radius of the metal sphere is  R = 2.0 \ cm  =  0.02 \ m

     The excess charge which the metal sphere carries is  q =  3.0 \mu C  =  3.0*10^{-6} \ C

      The distance of the position being to the center is D = 6.0 \ cm  = 0.06 \ m

       The coulomb constant is   k =9*10^{9} \  N \cdot m^2 /C^2

Generally the electric field is mathematically represented as  

        E = \frac{k *  q}{D^2}

substituting values

        E = \frac{9*10^{9} *  30.*10^{-6}}{(0.06)^2}

      E = 7.5 *10^{6} \ N/C

5 0
1 year ago
A moving 46.6 kg sled feels a 52.9 N friction force. what is the coefficient of friction
Setler [38]

Answer:

F=UR

52.9=U*46.6

U=52.9/46.6

U=1.135

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you are flying a kite that has two strings attached to either side of it you are holding those strings in your left and right ha
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The kite would move to the right as well
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