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kvasek [131]
3 years ago
13

A sample contains 20 kg of radioactive material. The decay constant of the material is 0.179 per second. If the amount of time t

hat has passed
is 300 seconds, how much of the of the original material is still radioactive? Show all work
Physics
1 answer:
Tju [1.3M]3 years ago
6 0

Answer:

There are 9.537\times 10^{-23} kilograms of radioactive material after 300 seconds.

Explanation:

From Physics we know that radioactive materials decay at exponential rate, whose differential equation is:

\frac{dm}{dt} = -\lambda\cdot m (1)

Where:

\frac{dm}{dt} - Rate of change of the mass of the radioactive material, measured in kilograms per second.

m - Current mass of the radioactive material, measured in kilograms.

\lambda - Decay constant, measured in \frac{1}{s}.

The solution of the differential equation is:

m(t) = m_{o}\cdot e^{-\lambda\cdot t} (2)

Where:

m_{o} - Initial mass of the radioactive material, measured in kilograms.

t - Time, measured in seconds.

If we know that m_{o} = 20\,kg, \lambda = 0.179\,\frac{1}{s} and t = 300\,s, then the initial mass of the radioactive material is:

m(t) = (20\,kg)\cdot e^{-\left(0.179\,\frac{1}{s} \right)\cdot (300\,s)}

m(t) \approx 9.537\times 10^{-23}\,kg

There are 9.537\times 10^{-23} kilograms of radioactive material after 300 seconds.

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Oxygen cycle

Explanation:

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From the average distance of one of jupiter's moons to jupiter and its orbital period around jupiter, we can determine:
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Explanation:

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T^2=\frac{4\pi^2}{GM}a^3

where T is the orbital period of a satellite, a is the average distance of the satellite from the Planet, M is the mass of the planet, G is the gravitational constant.

If the average distance of one of Jupiter's moons to Jupiter and its orbital period around Jupiter is given then mass of the Jupiter can be found:

\Rightarrow M_J=\frac{4\pi^2}{GT_m^2}a_m^3

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A protein molecule in an electrophoresis gel has a negative charge. The exact charge depends on the pH of the solution, but 30 e
Reika [66]

Answer:

7.401 * 10^(-15) N

Explanation:

30 electrons will have a charge:

30 * -1.6022 * 10^(-19) C

= - 4.806 * 10^(-18) C

The relationship between electric field and electric force is:

E = F/q

This means that force, F, is

|F| = |E|*|q|

|F| = |1540| * |-4.806 * 10^(-18)|

|F| = |-7401.24 * 10^(-18)|

|F| = 7.401 * 10^(-15) N

7 0
3 years ago
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s

8 0
4 years ago
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