A 310-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,190 A. If the conductor is copper
1 answer:
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Answer:
he mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.
Explanation:
A radioactive material is transformed into another material by the emission of some particular radioactive ones, the most common being alpha and beta rays, which is why in the transformation process a certain amount of mass is lost. The process is described by the expression
N = No
From this expression the quantity half life time () is defined with time so that half of the atoms have been transformed
T_{1/2} = ln 2 /λ
in this case it does not indicate that T_{1/2}= 20 days is worth, for which periods have passed, in the first the number of radioactive atoms was reduced to half the number, leaving N´ and the second halved the number of nuclei that they were radioactive, leaving radioactive nuclei
first time of life
N´ = ½ N
second time of life
N´´ = ½ N´
N´´ = ¼ N
consequently in the sample at the end of these two decay periods we have, assuming that after each emission the atom is stable (non-radioactive). After the first emission there are n₁ = N / 2 stable atoms, after the second emission n₂ = ¼ N stable atoms are added and there are still n₃ = ¼ N radioactive atoms, so the total number of atoms is
n_total = n₁ + n₂ + n₃
Recall that the mass of the initial radioactive atoms is m₁, when transforming its mass of stable atoms is m₂ where
m₂ < m₁
therefore mass of
m_total = m₂ N / 2 + m₂ N / 4 + m₁ N / 4
m_total = m₂ ¾ N + m₁ ¼ N
m_total = N ( ¾ m₂ + ¼ m₁)
Since the mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.