Although the weight of the spears a different the gravitational forces are the same
Answer:
1694 days
Explanation:
In first-order kinetics, the rate is proportional to the amount.
dA/dt = kA
For first-order kinetics, the rate k can be found using the half-life:
t₁,₂ = (ln 2) / k
In other words, the half-life is inversely proportional with the rate.
At the lower temperature, the rate is reduced to a third, so the half-life increases by a factor of 3. Meaning that the new half-life is 170 × 3 = 510 days.
The "shelf life" is the time it takes to reduce the initial amount to 10%. We can solve for this using the half-life equation.
A = A₀ (½)^(t / t₁,₂)
A₀/10 = A₀ (½)^(t / 510)
1/10 = (½)^(t / 510)
ln(1/10) = (t / 510) ln(½)
ln(10) = (t / 510) ln(2)
ln(10) / ln(2) = t / 510
t = 510 ln(10) / ln(2)
t ≈ 1694
Gamma waves are on the far right
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
