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Allisa [31]
3 years ago
11

What temperature increase is necessary to increase the power radiated from an object by a factor of 8?

Physics
1 answer:
Aleks04 [339]3 years ago
7 0

To solve this problem it is necessary to apply the concepts related to the Power defined from the Stefan-Boltzmann equations.

The power can be determined as:

P = \sigma T^4

Making the relationship for two states we have to

\frac{P_1}{P_2} = \frac{T_1^4}{T_2^4}

Since the final power is 8 times the initial power then

P_2 = 8P_1

Substituting,

\frac{1}{8} = \frac{T_1^4}{T_2^4}

T_2 = T_1 8*(\frac{1}{4})

T_2 = 1,68T_1

The temperature increase would then be subject to

\Delta T = T_2-T_1

\Delta T = 1.68T_1 -T_1

\Delta T = 0.68T_1

The correct option is D, about 68%

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A heat engine has a maximum possible efficiency of 0.780. If it operates between a deep lake with a constant temperature of-24.8
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Answer : The temperature of the hot reservoir (in Kelvins) is 1128.18 K

Explanation :

Efficiency of carnot heat engine : It is the ratio of work done by the system to the system to the amount of heat transferred to the system at the higher temperature.

Formula used for efficiency of the heat engine.

\eta =1-\frac{T_c}{T_h}

where,

\eta = efficiency = 0.780

T_h = Temperature of hot reservoir = ?

T_c = Temperature of cold reservoir = -24.8^oC=273+(-24.8)=248.2K

Now put all the given values in the above expression, we get:

\eta =1-\frac{T_c}{T_h}

0.780=1-\frac{248.2K}{T_h}

T_h=1128.18K

Therefore, the temperature of the hot reservoir (in Kelvins) is 1128.18 K

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The hottest climates on Earth are located near the Equator because this region
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<span>B. velocity .................</span>
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An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment
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Answer:

2.59×10¯²² Kgm/s

Explanation:

Data obtained from the question include:

Velocity of electron = 0.95c

Momentum =?

Next, we shall determine the velocity of the electron. This can be obtained as follow:

Velocity of electron = 0.95c

Velocity of Light (c) = 3×10⁸ m/s

Velocity of electron = 0.95c

Velocity of electron = 0.95 × 3×10⁸

Velocity of electron = 2.85×10⁸ m/s

Finally, we shall determine the mometum of the electron.

Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:

Momentum = mass x Velocity

Thus, with the above formula, we calculate the momentum of the electron as follow:

Mass of electron = 9.1×10¯³¹ Kg

Velocity of electron = 2.85×10⁸ m/s

Momentum of electron =?

Momentum = mass x Velocity

Momentum = 9.1×10¯³¹ × 2.85×10⁸

Momentum = 2.59×10¯²² Kgm/s

Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s

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