2 years ago

If a 4000 gram sample has a half-life of 1 million years and has 500 grams of radioactive parent isotope left how old is it

Physics

1 answer:

Arisa [49]2 years ago

4 0

3 million years

Happy to help

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calculate the force between two objects that have masses of 20 kg and 100 kg separated by a distance of 2.6 m

olga55 [171]

The gravitational force between the objects is $1.97\cdot 10^{-8} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between the objects

For the two objects in this problem:

$m_1 = 20 kg\\m_2 = 100 kg$

And their distance is

r = 2.6 m

So, the gravitational force between them is

$F=\frac{(6.67\cdot 10^{-11})(20)(100)}{2.6^2}=1.97\cdot 10^{-8} N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

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A rocket experiences a constant force even as the amount of fuel in its fuel tanks decreases. What happens to the acceleration o

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The correct answer is D

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A player bounces a basketball on the floor, compressing it to 80.0 % of its original volume. The air (assume it is essentially N

son4ous [18]

Answer: 292.95 J

Explanation:

change in internal energy= Heat transfer - work done

ΔU =Q -PΔV

Here, Q = 0 as there is no heat transfer.

P =2.00 atm = 2.00 × 101235 Pa = 202470 Pa

ΔV = final volume - initial volume = 0.8 V -V = -0.2 V

where V is the initial volume.

Volume of a spherical ball, $V = \frac{4}{3}\pi r^3$

r = d/2 = 23.9 cm / 2 = 0.12 m

$V = \frac{4}{3}\times 3.14 \times (0.12m)^3= 7.23\times10^{-3}m^3$

$\DeltaU = -P\DeltaV = - 202470 Pa \times -0.2 \times 7.23\times10^{-3}m^3=292.95 J$

Hence, internal energy would change by 292.95 J.

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7 A.M
I’m pretty sure that’s it

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