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Zina [86]
3 years ago
12

The half-life of an anticancer drug is 170 days at 25 oC. Degradation follows the first-order kinetics. When the drug is stored

at 5 oC, the degradation rate decreases to one third (1/3) of the rate at 25 oC. What is the shelf life of the drug at 5 oC ?
Physics
1 answer:
julsineya [31]3 years ago
5 0

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

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A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

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The value v_2  =  4 \sqrt{10} \  m/s

Explanation:

From the question we are told that

   The  charge on the first sphere is  q_1  =  2\mu C  =  2*10^{-6} \  C

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     KE  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (20 )^2

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And  U is  the  potential  energy which is mathematically represented as

        U  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

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Generally the total energy possessed by when q_2 and  q_1 are separated by 0.4 \  m is mathematically represented

         Q_f =  KE_f + U_f

Here KE_f is  the kinetic energy which is mathematically represented as

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substituting value

     KE_f  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (v_2 )^2

     KE_f  =  7.50 *10^{ -4} (v_2 )^2

And  U_f is  the  potential  energy which is mathematically represented as

        U_f  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U_f  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.4 }

      U_f  =  0.36 \  J

From the law of energy conservation

     Q =  Q_f

So

    0.48 =  0.36 +(7.50 *10^{-4} v_2^2)

   v_2  =  4 \sqrt{10} \  m/s

     

   

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