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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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How to solve it? Three capacitors with capacities of 600 pF, 300 pF, 200 pF are connected in series. The 60 V voltage is applied
adell [148]

Answer:

1. Voltage across 600 pF is 10 V.

2. Voltage across 300 pF is 20 V.

3. Voltage across 200 pF is 30 V.

Explanation:

We'll begin by calculating the total capacitance of capacitor. This can be obtained as follow:

Capicitance 1 (C₁) = 600 pF

Capicitance 2 (C₂) = 300 pF

Capicitance 3 (C₃) = 200 pF

Total capacitance (Cₜ) =?

1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃

1/Cₜ = 1/600 + 1/300 + 1/200

1/Cₜ = 1 + 2 + 3 / 600

1/Cₜ = 6/600

1/Cₜ = 1/100

Cₜ = 100 pF

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Thus, 100 pF is equivalent to 1×10¯¹⁰ F.

Next, we shall determine the charge. This can be obtained as follow:

Voltage (V) = 60 V

Capicitance (C) = 1×10¯¹⁰ F

Charge (Q) =?

Q = CV

Q = 60 × 1×10¯¹⁰ F

Q = 6×10¯⁹ C

1. Determination of the voltage across 600 pF.

Capicitance 1 (C₁) = 600 pF = 6×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 1 (V₁) =?

Q = C₁V₁

6×10¯⁹ = 6×10¯¹⁰ × V₁

Divide both side by 6×10¯¹⁰

V₁ = 6×10¯⁹ / 6×10¯¹⁰

V₁ = 10 V

2. Determination of the voltage across 300 pF.

Capicitance 2 (C₂) = 300 pF = 3×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 2 (V₂) =?

Q = C₂V₂

6×10¯⁹ = 3×10¯¹⁰ × V₂

Divide both side by 3×10¯¹⁰

V₂ = 6×10¯⁹ / 3×10¯¹⁰

V₂ = 20 V

3. Determination of the voltage across 200 pF.

Capicitance 3 (C₃) = 200 pF = 2×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 3 (V₃) =?

Q = C₃V₃

6×10¯⁹ = 2×10¯¹⁰ × V₃

Divide both side by 2×10¯¹⁰

V₃ = 6×10¯⁹ / 2×10¯¹⁰

V₃ = 30 V

7 0
3 years ago
Study the analogy, and then answer the question that follows.
nordsb [41]
What kind of analogy is this?
A. synonyms
B. part to whole
C. degrees of intensity
<span>D. cause to effect
This is because a synonym refers to things like "normal" or  "regular" like your calling someone something.
</span><span /><span>
</span>
6 0
4 years ago
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saveliy_v [14]

Answer:

C

Explanation:

b is 55 miles per hour south  or 55mph[South]

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What indicates that an object has been subjected to an unbalanced force?
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The object moves
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a heat engine converts heat to mechanical energy.

electric generator

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