gissellebramwell
11.11.2021 •
Mathematics
The half -life of the radioactive element unobtanium-29 is 5 seconds. If 80 grams of unobtanium-29 are initially present, how many grams are present after 5 seconds? 10 seconds? 15 seconds? 20 seconds? 25 seconds?
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Ответ:
Step-by-step explanation:
Are u boy or girl?
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Ответ:
the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively
Step-by-step explanation:
the equation that governs the remaining mass m of unobtanium -41 after a time t is
m=m₀*2^(-t/T) , where t is in seconds ,m₀ represents initial mass and T=half-life
Therefore
a) for t=5 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-5 s/5 s) = 48 gr/2 = 24 gr
b) for t=10 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-10 s/5 s) = 48 gr/4 = 12 gr
b) for t=15 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-15 s/5 s) = 48 gr/9 = 5.33 gr
b) for t=20 s
m=48 gr*2^(-t/5 s) = 48 gr*2^(-20 s/5 s) = 48 gr/16 = 3 gr
thus the remaining mass of unobtanium -41 after a 5 seconds, 10 seconds, 15 seconds and 20 seconds is 24 gr, 12 gr , 5.33 gr and 3 gr respectively
Ответ:
B. y=5x-2
Step-by-step explanation:
Basically the larger the number in front of x, the steeper the slope.