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The radioactive element americium-241 has a half-life of 432 years. How many years will it take a 10 gram mass of americium-241

geniusboy [140]

<h2>Answer: 816 years</h2>

This problem can be solved using the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=2.7g$ is the final amount of the material

$A_{o}=10g$ is the initial amount of the material

$t$ is the time elapsed (the quantity we are asked to find)

$h=432y$ is the half life of americium-241

Knowing this, let's find $t$ from (1):

$2.7g=(10g).2^{\frac{-t}{432y}}$

$\frac{2.7g}{10g}=2^{\frac{-t}{432y}}$

$0.27g=2^{\frac{-t}{432y}}$

Applying natural logarithm in both sides:

$ln(0.27g)=ln(2^{\frac{-t}{432y}})$

$-1.309=-\frac{t}{432y}ln(2)$

$-t=\frac{(-1.309)(432y)}{0.693}$

Finally:

$t=816y$

3 0

3 years ago

Solve for 8 - 6r + 3 = 5

Oksanka [162]

Answer:

r = 1

Step-by-step explanation:

8 - 6r + 3 = 5

Combine like terms

11 - 6r = 5

Subtract 5 from both sides

6 - 6r = 0

Add 6r to both sides

6 = 6r

Divide both sides by 6

1 = r

5 0

3 years ago

The radius of a sphere is increasing at a rate of 3 mm/s. how fast is the volume increasing when the diameter is 100 mm?

Katena32 [7]

33 per seconds
1s/3mm x/100mm x=33.3 s

4 0

3 years ago

Solve the system of linear equations below. 2x + 3y = 2 x + 6y = 4 x........A.X=0, y =2/3..B..x =2, y = -2/3...C..x=4/3, y=2/9..

Bogdan [553]

I will solve your system by substitution.2x+3y=2;x+6y=4xStep: Solve2x+3y=2for x:2x+3y+−3y=2+−3y(Add -3y to both sides)2x=−3y+22x2=−3y+22(Divide both sides by 2)x=−32y+1Step: Substitute−32y+1forxinx+6y=4x:x+6y=4x−32y+1+6y=4(−32y+1)92y+1=−6y+4(Simplify both sides of the equation)92y+1+6y=−6y+4+6y(Add 6y to both sides)212y+1=4212y+1+−1=4+−1(Add -1 to both sides)212y=3212y212=3212(Divide both sides by 21/2)y=27Step: Substitute27foryinx=−32y+1:x=−32y+1x=−32(27)+1x=47(Simplify both sides of the equation)Answer:x=4/7 and y=2/7
is what i got sorry if it don't help

3 0

3 years ago

In a group of dogs and their owners, there are exactly 20 heads and 64 legs. How many dogs are in the group

Juliette [100K]

Answer:

If we assume that everyone in the group is an owner, we would have

$20*2=40$ legs.