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Anna007 [38]
2 years ago
8

If you calculate the thermal power radiated by typical objects at room temperature, you will find surprisingly large values, sev

eral kilowatts typically. For example, a square box that is 1 m on each side and painted black (therefore justifying an emissivity e near unity) emits 2.5 kW at a temperature of 20∘C. In reality the net thermal power emitted by such a box must be much smaller than this, or else the box would cool off quite quickly. Which of the following alternatives seems to explain this conundrum best?A. The box is black only in the visible spectrum; in the infrared (where it radiates) it is quite shiny and radiates little power.
B. The surrounding room is near the temperature of the box and radiates about 2.5 kW of thermal energy into the box.
C. Both of the first two factors contribute significantly.
D. Neither of the first two factors is the explanation.
Physics
1 answer:
OverLord2011 [107]2 years ago
5 0

Answer:

best explanation of this is sentence B

Explanation:

The radiation emission of the bodies is given by the expression

     P = σ A e T⁴

Where P is the power emitted in watts, σ is the Stefan-Boltzmann constant, A is the surface area of ​​the body, e is the emissivity for black body e = 1 and T is the absolute body temperature in degrees Kelvin.

When the values ​​are substituted the power is quite high 2.5 KW, but the medium surrounding the box also emits radiation

   T box ≈ T room

    P box ≈ P room

As the two powers are similar and the box can absorbed, since it has the ability to emit and absorb radiation, as the medium is also close of the temperature of the box, the amount emitted is very similar to that absorbed, so the net change in energy is very small.

   In the case that the box is much hotter or colder than the surrounding medium if there is a significant net transfer.

Consequently, the best explanation of this is sentence B

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Image that the radiation emitted by the nitrogen at a frequency of 8.88×1014 Hz is absorbed by an electron in a molecule of meth
kiruha [24]

Answer:

5.07\cdot 10^{-7} m (507 nm)

Explanation:

First of all let's calculate the energy of the photon absorbed by the electron, This is given by

E=hf

where

h is the Planck constant

f=8.88\cdot 10^{14} Hz is the frequency of the photon

Substituting,

E=(6.63\cdot 10^{-34}Js)(8.88\cdot 10^{14}Hz)=5.89\cdot 10^{-19} J

The energy of the second photon, the one emitted when the electron drops to the intermediate energy level, is 2/3 of this energy:

E'=\frac{2}{3}E=\frac{2}{3}(5.89\cdot 10^{-19} J)=3.92\cdot 10^{-19} J

The relationship between the energy of the photon and its wavelength \lambda is

E=\frac{hc}{\lambda}

where c is the speed of light. Solving for \lambda, we find the wavelength:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.92\cdot 10^{-19} J}=5.07\cdot 10^{-7} m

3 0
2 years ago
Read 2 more answers
A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g
Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

Q_{cube}=Q_{water} (1)

But Q is related with mass (m), specific heat (c) and changes in temperature (\varDelta T)in the next way:

Q=cm\varDelta T(2)

Using (2) on (1):

c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}

(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})

(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})

Because we have an equality and 0.385 < 4.186 then \varDelta T_{cooper}>\varDelta T_{waterer} to conserve the equality

4 0
2 years ago
What is the acceleration of a ball rolling down a ramp that starts from rest and travels 0.9 m in 3 s?
cupoosta [38]
Given:
u = 0, initial velocity
s 0.9 m, distance traveled.
t = 3 s, the time taken.

Let a =  the acceleration. Then
s = ut + (1/2)*a*t²
(0.9 m) = 0.5*(a m/s²)*(3 s)²
0.9 = 4.5a
a = 0.2 m/s²

Answer: 0.2 m/s²
3 0
2 years ago
PLEASE HELPPP 100 POINTS HURRY !!!!Which diagram best illustrates the magnetic field of a bar magnet? A bar magnet with a north
Serggg [28]

I think this is right I hope this is right for you

7 0
2 years ago
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To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Kryger [21]

Answer:

H = 10.05 m

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

so this is the speed at the top of flag pole

now we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

now the height of flag pole is given as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
2 years ago
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