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Nezavi [6.7K]
3 years ago
5

A small flashlight bulb provides a resistance of 3 Ω to the 300 mA current that runs through it. Determine the voltage of the ba

ttery in the flashlight.
Physics
1 answer:
Contact [7]3 years ago
3 0

V=IR

V= 0.3 x 3

V= 0.9V

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       F_{net} = F_{g} -F_{n} = m*a\\  F_{net} = 930 N - 830 N = 100 N = 95 Kg * a

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       a =\frac{F_{net} }{m} =\frac{100 N}{95 kg} =  1.05 m/s2

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