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3 years ago

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"In a bad late-night science fiction film, a villain is using a large collection of radioactive atoms as energy for a weapon to

threaten the good guys. The atoms have a half-life of 1 hour. The villain has 4 kilograms of the radioactive material now, and he needs a minimum of 1 kg. for his weapon to work. After how much time will the weapon no longer be a threat?"

1 answer:

Veronika [31]3 years ago

3 0

Answer:

t=1.4hours

Explanation:

The half life is 1hour

At t=0 he has a mass of 4kg

So he want it to be 1kg, so that his weapon can work.

Applying the exponential function of decay

M=Cexp(-kt)

Where,

M is the mass at any time

C is a constant of integration

k is the rate of decay

Given that it has an half life of 1 hours.

Then k is 1

At t =0 the mass is 4kg

Therefore

4=Cexp(0)

C=4

M=4exp(-kt)

Since rate of decay is 1, then k=1

M=4exp(-t)

We need to find t at M=1kg

1=4exp(-t)

1=4exp(-t)

1/4=exp(-t)

0.25=exp(-t)

Take In of both sides

In(0.25)=-t

-1.3863=-t

Then, t=1.386hour

Then it will take about 1.4 hours to get to 1kg.

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Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

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Given that:

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$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

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$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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