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3 years ago

15

Which disadvantage of nuclear power use poses the greatest threat to the environment?

2 answers:

Mekhanik [1.2K]3 years ago

6 0

The correct answer is The storage and management of radioactive wastes

Explanation:

In general, nuclear reactions (changes in the nucleus of an atom such as fission) release a lot of energy including a lot of heat. Moreover, this heat is used by humans to obtain electricity and other types of energy, which is known as a nuclear power. This type of power is considered positive because it does not emit carbon and it is quite efficient.

However, in most cases, it is a threat to the environment and living beings because storing and managing the wastes of this type of power is difficult and expensive. Indeed, dealing with the wastes of nuclear power requires complex infrastructure, and any accident or leaking leads to serious consequences from the death of those exposed to the wastes to permanent loss of diversity or changes in nearby areas.

anzhelika [568]3 years ago

5 0

Answer:

C. The storage and management of radioactive wastes

Explanation:

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The correct answer to this question is false

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3 years ago

A technician attaches one lead of a digital voltmeter to the ground terminal of the TP sensor and the other meter lead to the ne

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Explanation:

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3 years ago

Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate

masha68 [24]

Answer:

The heat loss per unit length is $\frac{Q}{L} = 2981 W/m$

Explanation:

From the question we are told that

The outer diameter of the pipe is $d = 104mm = \frac{104}{1000} = 0.104 m$

The thickness is $D = 2mm = \frac{2}{1000} = 0.002m$

The temperature of water is $T = 90^oC = 90 + 273 = 363K$

The outside air temperature is $T_a = -10^oC = -10 +273 = 263K$

The water side heat transfer coefficient is $z_1 = 300 W/ m^2 \cdot K$

The heat transfer coefficient is $z_2 = 20 W/m^2 \cdot K$

The heat lost per unit length is mathematically represented as

$\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}$

Substituting values

$\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}$

$\frac{Q}{L} = \frac{628}{0.2107}$

$\frac{Q}{L} = 2981 W/m$

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