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Anna007 [38]
3 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]3 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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Answer:

<em>The frequency changes by a factor of  0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

where f is the frequency

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If the mass of the system is increased by 14 means the new frequency becomes

f_{n} = \frac{1}{2\pi } \sqrt{\frac{k}{14m} }

simplifying, we have

f_{n} = \frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }

f_{n} = \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }

if we divide this final frequency by the original frequency, we'll have

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  ÷  \frac{1}{2\pi } \sqrt{\frac{k}{m} }

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  x  2\pi \sqrt{\frac{m}{k} }

==> 1/3.742 = <em>0.27</em>

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3 years ago
Question: An increased number of rod cells increase the ability to see, A. during the day B. at night C. underwater or D. colors
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B- at night

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