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zavuch27 [327]
3 years 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]3 years 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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B. False

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Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.75 keV from the co
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0.6764*10^-10m

Explanation:

Using E= hc/wavelength

(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV

So subtract the calculated energy from the given energy of scattered photons

9.11-0.75=18.36 keV

To find wavelength

Wavelength= hc/ E

[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m

8 0
3 years ago
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
bekas [8.4K]

Answer:

Explanation:

Given that,

Radius r = 15cm = 0.15m

Area of the circular loop can be determined using the formula for area of a circle

A = π r²

A = π × 0.15²

A = 0.0708 m²

Magnetic field B = 1.2T in positive z direction

B = 1.2 •k T.

If loop is remove from the field in the time interval

∆t = 2.3ms = 2.3×10^-3s

We want to find the average EMF and it is given as

ε = —∆Φ/∆t

The final flux is zero

Φf = 0

Where magnetic flux is given as

Φi = BA Cosθ

Where θ=0 since the area and the magnetic field point in the same direction.

Φi = BA Cos0

Φi = BA

Φi = 1.2 × 0.0708

Φi = 0.0848 Vs

Then, ε = —∆Φ/∆t

ε = —(Φf — Φi) / ∆t

ε = —(0-0.0848) / (2.3×10^-3)

ε = 0.0848 / (2.3×10^-3)

ε = 36.88 V

The EMF is 36.88 Volts

6 0
3 years ago
How many joules of work are done against a truck when a force of 50 N pushes it 1 kilometer away
ELEN [110]

Answer:

Work = 50,000 J

Explanation:

Work = force * distance

Given that,

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Work = 50 * 1000

Work = 50,000 J

8 0
2 years ago
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