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Sagot :
Sure, let's walk through the solution step-by-step.
### Step 1: Determine the Half-Life of the Goo
The scientist begins with 236 grams of radioactive goo. After 9 minutes, the amount of goo decays to 7.375 grams.
To find the half-life, [tex]\( T \)[/tex], we use the radioactive decay formula:
[tex]\[ G(t) = G_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} \][/tex]
where:
- [tex]\( G(t) \)[/tex] is the remaining amount of goo at time [tex]\( t \)[/tex],
- [tex]\( G_0 \)[/tex] is the initial amount of goo,
- [tex]\( T \)[/tex] is the half-life.
Given that [tex]\( G_0 = 236 \)[/tex] grams, [tex]\( G(9) = 7.375 \)[/tex] grams, and [tex]\( t = 9 \)[/tex] minutes, we substitute these values into the equation:
[tex]\[ 7.375 = 236 \left(\frac{1}{2}\right)^{\frac{9}{T}} \][/tex]
Taking the natural logarithm on both sides to solve for [tex]\( T \)[/tex]:
[tex]\[ \ln(7.375) = \ln(236 \left(\frac{1}{2}\right)^{\frac{9}{T}}) \][/tex]
This simplifies to:
[tex]\[ \ln(7.375) = \ln(236) + \frac{9}{T} \ln\left(\frac{1}{2}\right) \][/tex]
From this, we can isolate [tex]\( T \)[/tex]:
[tex]\[ \frac{9}{T} \ln\left(\frac{1}{2}\right) = \ln(7.375) - \ln(236) \][/tex]
Solving for [tex]\( T \)[/tex]:
[tex]\[ T = \frac{9 \ln\left(\frac{1}{2}\right)}{\ln(7.375) - \ln(236)} \][/tex]
Given our constants:
- [tex]\(\ln(7.375) \approx 1.9981\)[/tex],
- [tex]\(\ln(236) \approx 5.4638\)[/tex],
- [tex]\(\ln\left(\frac{1}{2}\right) \approx -0.6931\)[/tex],
So:
[tex]\[ T = \frac{9 \times (-0.6931)}{1.9981 - 5.4638} \approx \frac{9 \times (-0.6931)}{-3.4657} \][/tex]
[tex]\[ T \approx 1.7999 \][/tex]
Therefore, the half-life of the goo is approximately [tex]\( 1.80 \)[/tex] minutes.
### Step 2: Derive the Formula for [tex]\( G(t) \)[/tex]
Using the computed half-life [tex]\( T \approx 1.80 \)[/tex], we can write the decay formula for the remaining amount of goo at time [tex]\( t \)[/tex]:
[tex]\[ G(t) = 236 \left(\frac{1}{2}\right)^{\frac{t}{1.80}} \][/tex]
### Step 3: Compute the Amount of Goo Remaining After 8 Minutes
To find the remaining amount of goo after 8 minutes, substitute [tex]\( t = 8 \)[/tex] into the decay formula:
[tex]\[ G(8) = 236 \left(\frac{1}{2}\right)^{\frac{8}{1.80}} \][/tex]
Simplify the exponent:
[tex]\[ G(8) = 236 \left(\frac{1}{2}\right)^{4.4444} \][/tex]
Using our calculations:
[tex]\[ G(8) \approx 10.8393 \][/tex]
So, the amount of goo remaining after 8 minutes is approximately [tex]\( 10.84 \)[/tex] grams.
In summary:
1. The half-life of the goo is approximately [tex]\( \boxed{1.80} \)[/tex] minutes.
2. The formula for [tex]\( G(t) \)[/tex] is [tex]\( \boxed{G(t) = 236 \left(\frac{1}{2}\right)^{\frac{t}{1.80}}} \)[/tex].
3. The amount of goo remaining after 8 minutes is approximately [tex]\( \boxed{10.84} \)[/tex] grams.
### Step 1: Determine the Half-Life of the Goo
The scientist begins with 236 grams of radioactive goo. After 9 minutes, the amount of goo decays to 7.375 grams.
To find the half-life, [tex]\( T \)[/tex], we use the radioactive decay formula:
[tex]\[ G(t) = G_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} \][/tex]
where:
- [tex]\( G(t) \)[/tex] is the remaining amount of goo at time [tex]\( t \)[/tex],
- [tex]\( G_0 \)[/tex] is the initial amount of goo,
- [tex]\( T \)[/tex] is the half-life.
Given that [tex]\( G_0 = 236 \)[/tex] grams, [tex]\( G(9) = 7.375 \)[/tex] grams, and [tex]\( t = 9 \)[/tex] minutes, we substitute these values into the equation:
[tex]\[ 7.375 = 236 \left(\frac{1}{2}\right)^{\frac{9}{T}} \][/tex]
Taking the natural logarithm on both sides to solve for [tex]\( T \)[/tex]:
[tex]\[ \ln(7.375) = \ln(236 \left(\frac{1}{2}\right)^{\frac{9}{T}}) \][/tex]
This simplifies to:
[tex]\[ \ln(7.375) = \ln(236) + \frac{9}{T} \ln\left(\frac{1}{2}\right) \][/tex]
From this, we can isolate [tex]\( T \)[/tex]:
[tex]\[ \frac{9}{T} \ln\left(\frac{1}{2}\right) = \ln(7.375) - \ln(236) \][/tex]
Solving for [tex]\( T \)[/tex]:
[tex]\[ T = \frac{9 \ln\left(\frac{1}{2}\right)}{\ln(7.375) - \ln(236)} \][/tex]
Given our constants:
- [tex]\(\ln(7.375) \approx 1.9981\)[/tex],
- [tex]\(\ln(236) \approx 5.4638\)[/tex],
- [tex]\(\ln\left(\frac{1}{2}\right) \approx -0.6931\)[/tex],
So:
[tex]\[ T = \frac{9 \times (-0.6931)}{1.9981 - 5.4638} \approx \frac{9 \times (-0.6931)}{-3.4657} \][/tex]
[tex]\[ T \approx 1.7999 \][/tex]
Therefore, the half-life of the goo is approximately [tex]\( 1.80 \)[/tex] minutes.
### Step 2: Derive the Formula for [tex]\( G(t) \)[/tex]
Using the computed half-life [tex]\( T \approx 1.80 \)[/tex], we can write the decay formula for the remaining amount of goo at time [tex]\( t \)[/tex]:
[tex]\[ G(t) = 236 \left(\frac{1}{2}\right)^{\frac{t}{1.80}} \][/tex]
### Step 3: Compute the Amount of Goo Remaining After 8 Minutes
To find the remaining amount of goo after 8 minutes, substitute [tex]\( t = 8 \)[/tex] into the decay formula:
[tex]\[ G(8) = 236 \left(\frac{1}{2}\right)^{\frac{8}{1.80}} \][/tex]
Simplify the exponent:
[tex]\[ G(8) = 236 \left(\frac{1}{2}\right)^{4.4444} \][/tex]
Using our calculations:
[tex]\[ G(8) \approx 10.8393 \][/tex]
So, the amount of goo remaining after 8 minutes is approximately [tex]\( 10.84 \)[/tex] grams.
In summary:
1. The half-life of the goo is approximately [tex]\( \boxed{1.80} \)[/tex] minutes.
2. The formula for [tex]\( G(t) \)[/tex] is [tex]\( \boxed{G(t) = 236 \left(\frac{1}{2}\right)^{\frac{t}{1.80}}} \)[/tex].
3. The amount of goo remaining after 8 minutes is approximately [tex]\( \boxed{10.84} \)[/tex] grams.
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