How many enery will be released in joule due to the mass defect in 1 gram?
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it is 1joule×igram this is the answer
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1 joule = 1N ×1m
= ma ×s
= 1kg ×accl^n×m
if mass reduced by 1 gram
= 999 /1000 g×accl^n×m
= o.999 kg × accl^n ×m
= ma ×s
= 1kg ×accl^n×m
if mass reduced by 1 gram
= 999 /1000 g×accl^n×m
= o.999 kg × accl^n ×m
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