Explore IDNLearn.com to discover insightful answers from experts and enthusiasts alike. Our platform offers reliable and comprehensive answers to help you make informed decisions quickly and easily.
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.
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.
