If the population is doubling every 10 years, the rate of change is 100%, for the period of 10 years, so, whatever it happens to be at the time, it grows by 100%, namely it doubles.
![\bf \textit{Periodic Exponential Growth}\\\\ A=I(1 + r)^{\frac{t}{p}}\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\to &4100\\ r=rate\to 100\%\to \frac{100}{100}\to &1.00\\ t=\textit{elapsed time}\to &8\\ p=period\to &10 \end{cases} \\\\\\ A=4100(1 + 1)^{\frac{8}{10}}\implies A=4100(2)^{\frac{4}{5}}\implies A=4100\sqrt[5]{2^4}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BPeriodic%20Exponential%20Growth%7D%5C%5C%5C%5C%0AA%3DI%281%20%2B%20r%29%5E%7B%5Cfrac%7Bt%7D%7Bp%7D%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%0AI%3D%5Ctextit%7Binitial%20amount%7D%5Cto%20%264100%5C%5C%0Ar%3Drate%5Cto%20100%5C%25%5Cto%20%5Cfrac%7B100%7D%7B100%7D%5Cto%20%261.00%5C%5C%0At%3D%5Ctextit%7Belapsed%20time%7D%5Cto%20%268%5C%5C%0Ap%3Dperiod%5Cto%20%2610%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D4100%281%20%2B%201%29%5E%7B%5Cfrac%7B8%7D%7B10%7D%7D%5Cimplies%20A%3D4100%282%29%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%5Cimplies%20A%3D4100%5Csqrt%5B5%5D%7B2%5E4%7D)
Factorise the following questions
1 answer:
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I have attached solution. be careful with your notation because at first I could not figure out because I thought they were 2 separate terms. in fact I believe it is all over the 3×10^7
:)
:)
Answer:
58 seconds
Step-by-step explanation:
Given:
Initial mass of water = 1 kg
Specific energy = 2297 kJ/kg
Heat supplied by the stove = 20 kJ/s
Now,
Half water is to be vaporized i.e 0.5 kg
Thus, heat required for vaporizing 0.5 kg water = mass × specific heat
or
heat required for vaporizing 0.5 kg water = 0.5 × 2297 = 1148.5 kJ
Therefore,
time taken to provide the required heat =
or
time taken to provide the required heat =
or
time taken to provide the required heat = 57.425 ≈ 58 seconds