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A highly volatile substance has an initial mass of 1200 g and its mass is reduced by 12% each second.

a Write a formula that gives the mass of the substance (m) at time (t) seconds.

b Rearrange this formula to make t the subject.

C What mass remains after 10 seconds, correct to 2 decimal places?

d Calculate how long (to the nearest second) it takes until the mass is 10 grams.

e After how many seconds (to the nearest second) is the mass less than 1 gram?​

Answer :

Answer:

Explanation:

a) 1.00 - 0.12 = 0.88

m = 1200(0.88)^t

b) t = ln(m/1200) / ln(0.88)

c) m = 1200(0.88)^10 = 334.20 g

d) t = ln(10/1200) / ln(0.88) = 37.451... = 37 s

e) t = ln(1/1200) / ln(0.88) = 55.463... = 55 s

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