Ask question

Ask question

All categories

2 years ago

10

A certain radioactive isotope has leaked into a small stream. Three hundred days after the leak, 3% of the original amount of t

he substance remained. Determine the half-life of this radioactive isotope.

1 answer:

sesenic [268]2 years ago

6 0

If the half-life is t, then every t days, the amount of the radioactive isotope will be cut in half.
(1/2)^(number of half-lives) = 3%
number of half-lives = ln(0.03) / ln(0.5)
This gives the number of half-lives as 5.06.
Then 300 days = (5.06)(length of 1 half-life)
length of 1 half-life = 300 / 5.06 = 59.29 days

You might be interested in

I need help with this plzz

Schach [20]

Answer:

5/2

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

2 years ago

Miss Garner has 69 gold stickers. She wants to give the same number of stickers to each of her 15 students. If she gives away as

Hatshy [7]

Answer: 69/15=4 and remainder 9

Step-by-step explanation: Since 15 x 4 = 60, 15 can't go into 9 so 9 in the leftover

7 0

2 years ago

Read 2 more answers

In 1962, the average cost of 1 gallon of gasoline in the United States was $0.31. The cost of gasoline increased by an average o

Kobotan [32]

Answer:

3*10=30.

Step-by-step explanation:

increased by 3% each year for 10 years

increased 30%.

0.31*0.30= .92

7 0

2 years ago

Correct I’ll give brainlist no links or I’ll report

padilas [110]

Answer:

None of these answers

Step-by-step explanation:

Plug the points into the equation to see if they make sense

(4,-10), -10=3(4)+2, -10=12+2, -10=14. Incorrect

(10,-4). -4=3(10)+2, -4=30+2, -4=32. Incorrect

(-10,-4), -4=3(-10)+2, -4=-30+2, -4=-28. Incorrect

Therefore, the answer is none of these answers

If this helps please mark as brainliest

5 0

2 years ago