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3 years ago

If the sphere shown above has a radius of 10 units then what is the approximate volume of the sphere

Mathematics

2 answers:

mestny [16]3 years ago

8 0

Answer: 4188.79

Step-by-step explanation:

Volume of the sphere is calculated by

V=4/3πr3

Where π is a constant, approximately 3.142

r = radius

From the question, given r = 10

Volume = 4/3 × 3.142 x 10 ^3

= 1.33333 x 3.142 x 1000

Volume =4188.79

Sophie [7]3 years ago

3 0

Answer: 4,188.79 u^3

Step-by-step explanation: Volume of a sphere is (4/3)pi•r^3

10^3 = 1000

(4/3) • pi • 1000 = 4,188.79 (rounded to 2 decimals)

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Need answer thanks! only answer if you know please!

Vaselesa [24]

Answer:

a) $k \approx 0.0602\,\frac{1}{min}$ ( $k\approx 6.02\,\frac{\%}{min}$ ), b) $k \approx 0.0648\,\frac{1}{min}$ ( $k \approx 6.48\,\frac{\%}{min}$ ), c) $t_{1/2} \approx 60.274\,days$ , d) $t_{1/2}\approx 4.305\,min$ , e) $k \approx 0.3\,\frac{1}{min}$ ( $k\approx 30\,\frac{\%}{min}$ ), f) $k \approx 0.0120\,\frac{1}{min}$ ( $k \approx 1.20\,\frac{\%}{min}$ ), g) $k \approx 2.876\times 10^{-5}\,\frac{1}{yr}$ ( $k \approx 2.876\times 10^{-3}\,\frac{\%}{yr}$ )

Step-by-step explanation:

All radioactive isotopes decay exponentially, the mass of the isotope as a function of time ( $m(t)$ ), measured in grams, is defined by:

$m(t) = m_{o}\cdot e^{-k\cdot t}$ (1)

Where:

$m_{o}$ - Initial mass of the isotope, measured in grams.

$k$ - Decay rate, measured in $\frac{1}{min}$ or $\frac{1}{day}$ or $\frac{1}{yr}$ .

$t$ - Time, measured in minutes, days or years.

We define the decay rate in terms of half-life by using the following expression:

$k = \frac{\ln 2}{t_{1/2}}$ (2)

Where $t_{1/2}$ is the half-life of the isotope, measured in minutes or years.

Now we proceed to determine the missing values:

a) Polonium-200 ( $t_{1/2} = 11.5\,min$ )

$k = \frac{\ln 2}{11.5\,min}$

$k \approx 0.0602\,\frac{1}{min}$

b) Lead-194 ( $t_{1/2} = 10.7\,min$ )

$k = \frac{\ln 2}{10.7\,min}$

$k \approx 0.0648\,\frac{1}{min}$

c) Iodine-125 ( $k = 0.0115\,\frac{1}{day}$ )

$t_{1/2} = \frac{\ln 2}{k}$

$t_{1/2} = \frac{\ln 2}{0.0115\,\frac{1}{day} }$

$t_{1/2} \approx 60.274\,days$

d) Kryption-75 ( $k = 0.161\,\frac{1}{min}$ )

$t_{1/2} = \frac{\ln 2}{k}$

$t_{1/2} = \frac{\ln 2}{0.161\,\frac{1}{min} }$

$t_{1/2}\approx 4.305\,min$

e) Strontium-79 ( $t_{1/2} = 2.3\,min$ )

$k = \frac{\ln 2}{2.3\,min}$

$k \approx 0.3\,\frac{1}{min}$

f) Uranium-229 ( $t_{1/2} = 58\,min$ )

$k = \frac{\ln 2}{58\,min}$

$k \approx 0.0120\,\frac{1}{min}$

g) Plutonium-239 ( $t_{1/2} = 24100\,yr$ )

$k = \frac{\ln 2}{24100\,yr}$

$k \approx 2.876\times 10^{-5}\,\frac{1}{yr}$

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