Answer:
Explanation:
The half life of a radioactive substance is the time in which the activity of the radioactive substance becomes half of its initial activity.
Or
The time in which the amount of radioactive substance becomes half of its initial amount in a sample.
It is denoted by T .
The relation between the decay constant and the half life is given by
Answer:
3. When the number of turns, N is doubled, the strength of the electromagnet is also doubled
4. Doubling the voltage, doubles the strength of the electromagnet
5. The number of paper clips a 7.5 V battery would pick is approximately 28 paper clips
The number of paper clips a 7.5 V battery would pick is 59 paperclips
6. For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is approximately 7 paperclips
For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is 16 paperclips
Explanation:
3. The Magnetomotive Force, MMF = The Number of Turns on the Coil, N × The Current I Flowing in the Coil, I
∴ MMF = N × I
When the number of turns, N is doubled, the magnetomotive force, MMF is also doubled, and the strength of the electromagnet is doubled
4. Given that the voltage, V applied to the coil = The current, I flowing × The resistance, R of the coil, we have
V = I × R
Therefore, for a given constant resistance in the coil, doubling the voltage, doubles the current and therefore doubles the strength of the electromagnet
5. The average slope for the 25-coil electromagnet = (23 - 12)/(6 - 3) = 3.
The number of paper clips a 7.5 V battery would pick = 12 + (7.5 - 3) × 11/3 = 28.5 paperclips ≈ 28 paper clips
The average slope for the 50-coil electromagnet = (48 - 26)/(6 - 3) = 7.
The number of paper clips a 7.5 V battery would pick = 26 + (7.5 - 3) × 22/3 = 59 paperclips
6. The slope calculated from a start point of approximately 0.4 V, is given as follows;
The slope for the 25-coil electromagnet = (12 - 6)/(3 - 0.4) = 30/13
Therefore, for the 25-coil electromagnet, the average number of paper clips a 1 V battery would pick = 6 + (1 - 0.4) × 30/13) = 96/13 ≈ 7 paperclips
The slope for the 50-coil electromagnet = (26 - 13)/(3 - 0.4) = 5
Therefore, for the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick = 13 + (1 - 0.4) × 5 = 16 paperclips