Final answer:
Using Ohm's law, the smallest voltage that poses a danger of ventricular fibrillation is 6 mV, necessitating special electrical safety precautions. A defibrillator should apply a voltage of 5 V. During defibrillation, 0.12 C of charge is transferred, corresponding to approximately 7.5 x 10^17 electrons.
Step-by-step explanation:
Calculating Voltage and Charge in Medical Procedures
For question 90(a), we apply Ohm's law (V=IR) to find the smallest voltage that poses a danger of causing ventricular fibrillation during surgery. Given a current of 20.0 µA (which is 20.0 x 10-6 A) and a resistance of 300Ω, the smallest dangerous voltage would be:
V = I x R = 20.0 x 10-6 A x 300Ω = 0.006 V or 6 mV. This implies that very special electrical safety precautions are indeed necessary to prevent such low voltages from affecting the heart.
In question 8, where a defibrillator is used, we need to apply a certain voltage to achieve a 10.0-mA (10.0 x 10-3 A) current through a resistance of 500Ω. Again, using Ohm's law:
V = I x R = 10.0 x 10-3 A x 500Ω = 5 V. A defibrillator should apply a voltage of 5 V.
As for question 9(a), the charge (Q) moved by a current (I) over time (t) is calculated using the formula Q = I x t. So, for a current of 12.0 A over 0.0100 s, the charge transferred is:
Q = 12.0 A x 0.0100 s = 0.12 C. To find the number of electrons (part 9(b)), we use the charge of a single electron (1.602 x 10-19 C) and divide the total charge by this value:
Number of electrons = Total charge / Charge per electron = 0.12 C / 1.602 x 10-19 C/electron = approximately 7.5 x 1017 electrons.