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telo118 [61]
3 years ago
11

A circular grill of diameter 0.25 m has an emissivity of 0.8. If the surface temperature is maintained at 150°C, determine the r

equired electrical power when the room air and surroundings are at 30°C.
Physics
1 answer:
Schach [20]3 years ago
4 0

Answer:

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

Explanation:

Power required to maintain the surface temperature at 150°C from 20°C

P= εσA(T^4-t^4)

P= power in watt

ε= emissivity

A=  area of surface

T= 150°C= 423 K

t= 20°C= 303K

/sigma= 5.67×10^{-8} watt/m^2K^4

putting vales we get

= 0.8\times5.67\times10^{-8} \pi\frac{0.25^4}{4}(423^4-303^4)

P=52.51822 Watt

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

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In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. (a) Determine the orbital frequency of
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Answer:

(a) 6.567 * 10^15 rev/s or hertz

(b) 8.21 * 10^14 rev/s or hertz

Explanation:

Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)

Where Fn is frequency at all levels of n.

Z = 1 (nucleus)

e = 1.6 * 10^-19c

m = 9.1 * 10^-31 kg

h = 6.62 * 10-34

K = 9 * 10^9 Nm2/c2

(a) for groundstate n = 1

Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s

(b) first excited state

n = 1

We multiple the groundstate answer by 1/n^3

6.567 * 10^15 rev/s/ 2^3

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3 years ago
What distance does light travel in water, glass, and diamond during the time that it travels 1.0 m in vacuum? The refractive ind
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Answer:

refractive index for water,glass,diamond are 0.752m, 0.667m, 0.413m respectively

Explanation:

refractive index (n) = \frac{velocity of light in air/vacuum}{velocity of light in substance}

velocity =\frac{distance}{time}

The time for travel is kept constant for all mediums.

refractive index (n) = \frac{\frac{distance in vacuum}{time} }{\frac{distance in medium}{time} }\\ \\=\frac{distance in vacuum}{distance in medium}

distance in medium = \frac{distance in vacuum}{refractive index of medium}

S_{medium}  = \frac{S_{vacuum} }{n_{medium} }

For water, n= 1.33

S_{water} = \frac{1 }{1.33}

S_{water} = 0.752m

For glass, n=1.5

S_{glass}= \frac{1 }{1.5}

S_{glass} = 0.667m

For diamond, n= 2.42

S_{diamond} = \frac{1 }{2.42}

S_{diamond} = 0.413m

3 0
3 years ago
How much money would be saved by turning off one 100.0-W lightbulb 3.0 h/day for 365 days if the
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Answer:

the money that would be saved is $13.14.

Explanation:

Given;

power consumed by the light bulb, P = 100 W = 0.1 kW

time of running the bulb, t = 3 hours for 365 days = 1,095 hours

cost rate of power consumption, C = $0.12 per kWh

Energy consumed by the light bulb for the given days;

E = Pt

E = 0.1 kW  x 1,095 hr

E = 109.5 kWh

Cost of energy consumed = 109.5 kWh   x   $0.12 / kWh

                                            = $13.14

Therefore, the money that would be saved is $13.14.

3 0
3 years ago
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