What traction of the radioisotoperemains in the body after one day?
1 answer:
The fraction of radioisotope left after 1 day is , with the half-life expressed in days
Explanation:
The question is incomplete: however, we can still answer as follows.
The mass of a radioactive sample after a time t is given by the equation:
where:
is the mass of the radioactive sample at t = 0
is the half-life of the sample
This means that the mass of the sample halves after one half-life.
We can rewrite the equation as
And the term on the left represents the fraction of the radioisotope left after a certain time t.
Therefore, after t = 1 days, the fraction of radioisotope left in the body is
where the half-life must be expressed in days in order to match the units.
Learn more about radioactive decay:
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Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;
Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
<h2>5.3 km</h2>
Explanation:
This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.
Let us denote each of the individual displacements by a vector. Consider the unit vectors as the unit vectors in the direction of East and North respectively.
By simple calculations, we can derive the unit vectors in the directions North,
South of West and
North of West respectively.
So Total displacement vector = Sum of individual displacement vectors.
Displacement vector =
Magnitude of Displacement =
∴ Total displacement =