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Gelneren [198K]
3 years ago
8

I've just discovered a new radioactive element. At 1:00 pm I measure 10 grams of the element, but at 1:24 pm I measure only 1.25

grams of the element (with 8.75 grams of daughter product). How long is each half life, in minutes? (Just enter the number).
Physics
1 answer:
docker41 [41]3 years ago
6 0

Answer: Each half life is 8 minutes long.

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample  = 24 minutes

a = initial amount of the reactant  = 10 grams

a - x = amount left after decay process  = 1.25 grams

24min=\frac{2.303}{k}\log\frac{10}{1.25}

k=\frac{2.303}{24}\log\frac{10}{1.25}

k=0.0866min^{-1}

for completion of half life:  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.0866min^{-1}}=8.00min

Thus half life is 8 minutes long

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4 years ago
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A ring of diameter 7.70 cm is fixed in place and carries a charge of 5.00 mC uniformly spread over its circumference. (a) How mu
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Answer:

3.974 Joule

Explanation:

Diameter of ring = 7.7 cm

a = Distance from the center = d/2 = 3.85 cm = 0.0385 m

Q = Charge = 5 mC

q = Charge to move = 3.4 mC

k = Coulomb constant = 9×10⁹ Nm²/C²

Work done will be equal to Potential energy when mass is at center

U=\frac{kQq}{a}\\\Rightarrow U=\frac{9\times 10^9\times 5\times 10^{-6}\times 3.4\times 10^{-6}}{0.0385}=3.974\ J

∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule

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Alannah added a strip of zinc to a beaker containing a solution of copper
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The solution of copper sulfate was too cold

Explanation:

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An 18-cm-long bicycle crank arm, with a pedal at one end, is attached to a 20-cm-diameter sprocket, the toothed disk around whic
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Answer:

Part a)

a = 0.056 m/s^2

Part b)

L = 7.85 m

Explanation:

Part a)

Angular speed of the pedal is changing from 60 rpm to 90 rpm in 10 s

so the angular acceleration is given as

\alpha = \frac{\omega_2 - \omega_1}{\Delta t}

so we will have

\alpha = \frac{2\pi(\frac{90}{60}) - 2\pi(\frac{60}{60})}{10}

\alpha = 0.314 rad/s^2

now the tangential acceleration of the pedal is given as

a = r \alpha

a = 0.18 \times 0.314

a = 0.056 m/s^2

Part b)

Total angular displacement made by the sprocket in the interval of 10 s is given as

\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{2\pi (\frac{90}{60}) + 2\pi (\frac{60}{60})}{2}(10)

\theta = 78.5 radian

now length of the chain passing over it is given as

L = R\theta

L = 0.10 \times 78.5

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Answer:

moving slowly is the answer

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