Answer :
Final answer:
We need to calculate the value of ted ted ted ted ted ted ted (10%) for a heart defibrillator given the capacitance and potential difference. By multiplying the capacitance and potential difference, we can find the value of ted ted ted ted ted ted ted (10%).
Explanation:
In this question, we are given the capacitance and the potential difference of a heart defibrillator. The capacitance (C) is 9.75 µF and the potential difference (V) is 11 x 10¹⁰ V. We need to find the value of ted ted ted ted ted ted ted (10%).
By multiplying the capacitance and the potential difference, we can find the value of ted ted ted ted ted ted ted (10%):
ted ted ted ted ted ted ted (10%) = C * V
Substituting the given values, we have:
ted ted ted ted ted ted ted (10%) = 9.75 µF * 11 x 10¹⁰ V
Simplifying the expressions and converting the units, we get:
ted ted ted ted ted ted ted (10%) = 107.25 x 10⁻⁶ C * 11 x 10¹⁰ V
ted ted ted ted ted ted ted (10%) = 1179.75 x 10⁴ CV
Therefore, ted ted ted ted ted ted ted (10%) is approximately equal to 1179.75 x 10⁴ CV.
The given data for Problem 9 are:C = 9.75 µF
= 9.75 x 10^-6 FV
= 11 x 10^3 V
= 1.1 x 10^4 V
We can find the electric charge on the capacitor using the formula,Q = CVwhere Q is the electric charge, C is the capacitance and V is the potential difference
Q = (9.75 x 10^-6 F) x (1.1 x 10^4 V)
= 107.25 x 10^-2 C
= 1.0725 C
So, the main answer is: The electric charge on the capacitor is 1.0725 C. The explanation to the main answer is as follows:We can use the formula Q = CV to find the electric charge on the capacitor. After substituting the given values for capacitance and potential difference, we getQ = (9.75 x 10^-6 F) x (1.1 x 10^4 V) = 107.25 x 10^-2 C = 1.0725 CTherefore, the electric charge on the capacitor is 1.0725 C.
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