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serg [7]
3 years ago
9

Which three characteristics make a glowing piece of charcoal a blackbody radiator? A. The frequencies of EMR it emits depend on

its temperature. B. It emits only one frequency of EMR. C. It absorbs most of the EMR it receives. D. It emits a range of wavelengths of EMR.
Physics
1 answer:
Eva8 [605]3 years ago
4 0

Answer:

A. The frequencies of EMR it emits depend on its temperature

B. It emits only one frequency of EMR

C. It absorbs most of the EMR it receives

Explanation:

  • A blackbody is an object that absorbs most of the electromagnetic spectrum of energy that falls on it.
  • According to law to reradiates most of the available energy back on top the outer space at an efficiency of 100% and thus radiation may be in the visible range of temperature than are in 1000K.
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Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
3 years ago
A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.
denpristay [2]

Answer:

The mass of the rule is 56.41 g  

Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0                          22cm                                  100cm

-------------------------Δ------------------------------------

↓                                                                       ↓

200g                                                                 m₂  

Apply principle of moment

(200 g)(22 cm - 0)     = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ =  (200 g)(22 cm)  / (78 cm)

m₂ = 56.41 g  

Therefore,  the mass of the rule is 56.41 g                                            

3 0
3 years ago
The frequency of a wave is 560 Hz. What is it’s period
Crazy boy [7]
The answer would be, "1/560 seconds".
4 0
3 years ago
How did water get from the ocean to your water faucet?
Semmy [17]

Answer:

lol im pretty sure pipes and nice pic of lil darkie

Explanation:

2+2=4

3 0
3 years ago
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How do you think heat gets from a filament to the kernel?​
PilotLPTM [1.2K]

Answer:

Explanation:

It goes thru the fluma

8 0
3 years ago
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