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The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 {\rm mA} for 37 continuous hours. During that time the voltage will drop from 1.5 {\rm V} to 1.0 {\rm V} . Assume the drop in voltage is linear with time.Part A How much energy does the battery deliver in this 37h interval?Express your answer to two significant figures and include the appropriate units.

Respuesta :

Hashirriaz830

Answer:

1.42 KJ

Explanation:

solution:

power in beginning [tex]p_{0}[/tex]=(1.5 V).(9×[tex]10^{-3}[/tex] A)

                                    = 13.5 mW

after continuous 37 hours it drops to

                                    [tex]p_{37}[/tex]=(1 V).(9×[tex]10^{-3}[/tex] A)

                                         =9 mW

When the voltage will drop energy will not remain the same but the voltage drop will always remain same if the voltage was drop to for example from 5 V to 4.5 V the drop will remain the same.

                  37 hours= 37.60.60

                                 =133200‬ s

                              w=(9×[tex]10^{-3}[/tex] A×133200‬ )+[tex]\frac{1}{2}(13.5.10^{-3}-9.10^{-3})(133200)[/tex]

                                 =1.42 KJ

NOTE:

There maybe a calculation error but the method is correct.

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