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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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Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

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intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|

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\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}

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Answer:

A. potential energy is 258720 Joule

Explanation:

A.Gravitational potential energy is: PE = m × g × h

velocity =  15.33 m/s when the car reaches the bottom of the hill.

where, m = mass

            g = acceleration due to gravity

            h = height from the bottom of hill.

The potential energy is : m×g×h

                                     =(2200×9.8×12)

                                     =258720 Joule

B. at the bottom of the hill, the potential energy is converted into kinetic energy so PE at top = KE at bottom

                    kinetic energy= \frac{1}{2}(m*v^{2})

where v = velocity

          m= mass

therefore,               v=\sqrt\frac{2*K.E}{m} {}

                         or,  v=\sqrt{\frac{2*258720}{2200} }

                         or,   v=15.33 m/s

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