The half-life of the isotope Osmium-183 is 12 hours. Choose the equation below that gives the remaining mass of Osmium-183 in grams, M.
after n half-lives have elapsed if there was an initial mass of 590 grams before decay. Then, use the equation to determine the mass remaining
after 36 hours have passed
Mn = 590. ()": M3 – 148
D
M, = 590 C4)*-* : My ~ 148
Mn = 590 - ()": Mz - 74
Mi = 590 - (1) * + : Mz - 74

Answers

Answer 1

The given equations are incomprehensible, I'm afraid...

You're given that osmium-183 has a half-life of 12 hours, so for some initial mass M₀, the mass after 12 hours is half that:

1/2 M₀ = M₀ exp(12k)

for some decay constant k. Solve for this k :

1/2 = exp(12k)

ln(1/2) = 12k

k = 1/12 ln(1/2) = - ln(2)/12

Now for some starting mass M₀, the mass M remaining after time t is given by

M = M₀ exp(kt )

So if M₀ = 590 g and t = 36 h, plugging these into the equation with the previously determined value of k gives

M = 590 exp(36k) = 73.75

so 73.75 ≈ 74 g of Os-183 are left.

Alternatively, notice that the given time period of 36 hours is simply 3 times the half-life of 12 hours, so 1/2³ = 1/8 of the starting amount of Os-183 is left:

590/8 = 73.75 ≈ 74


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Brainliest pls??

Towns P, Q and R are connected by roads PQ ,PR and QR. PR is 10km longer than PQ
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Work out length of PQ.​

Answers

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PQ = 35 km

Step-by-step explanation:

Here, we want to get the length of PQ

Let PQ = x

PR is 10 km longer than PQ

PR = 10 + x

QR = 2 * PR

QR = 2(10 + x)

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The length of the road PQ of the town where, towns P, Q and R are connected by roads PQ ,PR and QR is 35 km.

What is the length of line segment?

Length of line segment (say YX) is the distance of both the ends of it (y to x).

Towns P, Q and R are connected by roads PQ ,PR and QR.

The line segment PR is 10 km longer than PQ. Therefore,

[tex]PR=PQ+10\\PQ=PR-10[/tex]            ......1

The line segment QR is twice as long as the line segment PR. Therefore,

[tex]QR=2PR[/tex]                 .....2

The total length of three roads is 170 km. Therefore,

[tex]PQ+PR+QR=170[/tex]

Put the value of PR and QR from equation 1 and 2 in the above equation as,

[tex](PR-10)+PR+(2PR)=170\\PR-10+PR+2PR=170\\4PR=170+10\\PR=\dfrac{180}{4}\\PR=45\rm km[/tex]

As the line segment PR is 10 km longer than PQ and length of PR is 45 km. Therefore,

[tex]PR=45-10\\PR=35\rm km[/tex]

Hence, the length of the road PQ of the town where, towns P, Q and R are connected by roads PQ ,PR and QR is 35 km.

Learn more about the line segment here;

https://brainly.com/question/2437195

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