High School

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------------------------------------------------ IS ted ted ted ted ted ted ted (10%) Problem 9: Suppose you have the 9.75 µF capacitor of a heart defibrillator at a potential difference of 11 x 10 V. Randomized Variables C=9.75 µF V= 11 x 10¹ V

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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