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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
Answer:
Explanation:
Given that,
The distance between two spheres, r = 25 cm = 0.25 m
The capacitance, C = 26 pF = 26×10⁻¹² F
Charge, Q = 12 nC = 12 × 10⁻⁹ C
We need to find the work done in moving the charge. We know that, work done is given by :
Put all the values,
So, the work done is .