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3241004551 [841]
3 years ago
10

The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 1% per day. A s

ample of this radioactive substance has an initial mass of 2.5kg. Find the mass of the sample after five days.
Physics
1 answer:
Bumek [7]3 years ago
6 0

Answer:

2,38kg

Explanation:

Mass in function of time can be found by the formula: m_{(t)} =m_{0} e^{-kt}, where m_{0} is the initial mass, t is the time and k is a constant.

Given that a sample decay 1% per day, that means that after first day you have 99% of mass.

m_{(1)} =m_{0} e^{-k(1)}, but m_{(1)}=\frac{99m_{0} }{100}, so we have \frac{99m_{0} }{100}=m_{0}e^{-k}, then k=-ln(\frac{99}{100})=0.01

Now using k found we must to find m_{(5)}.

m_{(5)}=m_{0}e^{-(0.01)5}=2.5kge^{-0.05} =2.5x0.951=2.38kg

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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
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Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

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Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

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w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

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⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

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T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

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