A heart defibrillator passes 12.5 A through a patient\'s torso for 4.95 ms in an

Question

A heart defibrillator passes 12.5 A through a patient's torso for 4.95 ms in an attempt to restore normal beating. Randomized Variables I = 12.5 A t = 4.95 ms E = 540 J m = 5.5 kg How much charge passed, in coulombs? Numeric ; A numeric value is expected and not an expression. Q =_ What voltage, in kilovolts, is applied if 540 J of energy was dissipated? Numeric ; A numeric value is expected and not an expression DeltaV = What is the path's resistance, in ohms? Numeric : A numeric value is expected and not an expression. R = Find the temperature increase, in degrees Celsius, caused in the 5.5 kg of affected tissue flesh has a specific heat of 3500 J/kg degree C Numeric : A numeric value is expected and not an expression. AT =

Explanation / Answer

A.) Q= It=12.5*4.95 =61.875 mC=6.19*10^-2 C

B.)Energy= VI

540=12.5 V

V=43.2 V

C.) R=V/I=43.2/12.5=3.47 Ohm

D.)Heat=m C T

540=5.5*3500T

T=0.03 C

Increase in temperature =0.03 C

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