What is the energy (in J) stored in the 23.0 µF capacitor of a heart defibrillator charged to 7.40 ✕ 103 V?

629.7 J = 0.63 kJ.

Explanation:

Charge and voltage are related to the capacitance C of a capacitor by Q = CV, and so the expression for Ecap can be written mathematically into three equivalent expressions:

Ecap = QV^2

= 1/2 * (C*V^2)

= 1/2 * Q^2/C

where,

Q = charge

V = voltage on a capacitor C.

The energy is in joules for a charge in coulombs, voltage in volts, and capacitance in farads.

Given:

C = 23.0 µF

= 2.3 x 10^-5 F

V = 7.40 ✕ 10^3 V

Ecap = 1/2 * (C*V^2)

= 1/2 * (2.3 x 10^-5) * (7.40 ✕ 10^3)^2

= 629.7 J

= 0.63 kJ.

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

i think the answer would be b