How do you solve this problem?

According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the

FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

Then, we obtain two differential equations

$\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q$

where $k_1$ and $k_2$ are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

$\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

Now, the variables are separated, $P$ and $Q$ appear only on the left, and $t$ appears only on the right, so that we can integrate both sides.

$\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

$\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$ ,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

$e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

$P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$ ,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$ .

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

$P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

$C = P(0) = C_1 e^0 \implies C= C_1$

Similarly, we obtain $C = C_2$ and

$P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}$

The relation between the decay constant $k$ and the half-life is given by

$\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$ . Thus,

$4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

$7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

$P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

$\frac{P(t)}{Q(t)} = 137.7$

Hence,

$\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7$

Solving for $t$ yields $t \approx 6 \times 10^9$ , which means that the age of the universe is about 6 billion years.

5 0

3 years ago

Tashia ate 1/3 of a pizza, and Jay ate 3/8 of the same pizza. What fraction of the pizza was eaten?

Sauron [17]

This is basically saying how much of the pizza did Tashia and Jay both ate, thus combining their portions together.

1/3 + 3/8 = fraction of pizza eaten

To add fractions, you must have a common denominator. Which is 24 in this case because 3*8 = 24.

So you multiply 1/3 with 8 and multiply 3/8 with 3.

1/3*8= 8/24
3/8*3= 9/24

Now you add the fractions.
8/24+9/24= 17/24. You add the numerator, the denominator stays the same.

17/24 of the pizza was eaten.

7 0

3 years ago

Clarke borrows $16000 to buy a car he pays simple interest at an annual rate of 6% over a period of 3.5 years how much does he p

uranmaximum [27]

Clarke borrows $16000 to buy a car he pays simple interest at an annual rate of 6% over a period of 3.5 years how much does he pay all together?
Let’s calculate
=> 16 000 dollars is the amount he borrowed
=> 6% is the annual interest
=> 3.5 years us the total years that he will be paying it.
Solutions:
=> 16 000 dollars * .06 = 960 dollars is the annual interests
=> 960 * 3.5 = 3 360 dollars for 3.5 years
=> 16 000 + 3 360 = 19 360 dollars

4 0

3 years ago

Evaluate the expression when n=6 n² – 5n-2

eduard

Answer:

4

Step-by-step explanation:

(6^2) -5(6) -2 =4

that is the answer

8 0

3 years ago

Read 2 more answers

PLEASE HELP

Ronch [10]

Answer:

1 and 5

Step-by-step explanation:

The squares have a side length of 10 and 1 square side is the radius of the half-circles. Since there are two half-circles, find the circumference for one full circle:

$C=2\pi r$

Insert the radius:

$r=10\\C=2\pi *10$

Simplify pi:

$C=2*3.14*10$

Simplify multiplication:

$C=6.28*10\\C=62.8$

The circumference of the circles is 62.8. Now find the perimeter of the exposed squares with side length 10. There are 4 exposed sides, which equals one square. Find the perimeter:

$P=10+10+10+10\\P=10+10+20\\P=10+30\\P=40$