Potassium 40 is an isotope with 19 protons (and electrons) and 21 neutrons. Potassium 40 accounts for around 0.012% of potassium and is fairly stable (half life of 1.25 billion years).
Potassium 40 is an isotope with 19 protons (and electrons) and 21 neutrons. Potassium 40 accounts for around 0.012% of potassium and is fairly stable (half life of 1.25 billion years).
Potassium 40 is an isotope with 19 protons (and electrons) and 21 neutrons. Potassium 40 accounts for around 0.012% of potassium and is fairly stable (half life of 1.25 billion years).
Potassium 40 is an isotope with 19 protons (and electrons) and 21 neutrons. Potassium 40 accounts for around 0.012% of potassium and is fairly stable (half life of 1.25 billion years).
The height would remain the same.
Technology helps our lives because if we didn't have technology, we would have unexpected weather. And what if it is severe weather? We would all die!
It would be faster to answer how it does not impact our lives. That is to say it plays a role in almost every single facet of modern life.
Would need overall height or id to get answer
The brightness would remain constant but the power draw will increase. If the circuit was series wired, the brightness would go down as you added bulbs.
After each half-life, the number of undecayed nuclei is halved. Starting with 600 nuclei, after one half-life, 300 would remain; after the second half-life, 150 would remain; and after the third half-life, 75 would remain. Thus, after three half-lives, 75 undecayed headsium nuclei would remain in the sample.
75
Not sure what you mean by "had-lives". After 3 half lives, approx 1/8 would remain.
After one half-life, half of the original amount of Uranium-235 would remain. After four half-lives, only ( \frac{1}{2^4} ) or ( \frac{1}{16} ) of the original amount would be left. Therefore, if you started with 100 grams of Uranium-235, 6.25 grams would remain after four half-lives.
After 3 half-lives, half of the original sample would remain unchanged. After the 1st half-life: 300 unchanged atoms. After the 2nd half-life: 150 unchanged atoms. After the 3rd half-life: 75 unchanged atoms would remain.
After 8.1 days, three half-lives have passed for Au-198 (2.7 days/half-life * 3 half-lives = 8.1 days). Each half-life reduces the number of atoms by half, so remaining atoms = 800 atoms * (1/2)^3 = 100 atoms.
If the half-life of bromine is 25 minutes, than in 75 minutes three half lives would have passed. If you started with a 4.0 mg sample, than after three half-lives only 0.50 mg would remain. Because you are dealing with a whole number of half lives, all you have to do is divide 4.0 by 2 three time. Alternatively, you could multiply 4.0 by one half raised to the third power. This would look like this when entered into a calculator: 4.0 X (0.5^3) = 0.5
After 5 half-lives, 3.125% of the original carbon-14 would remain. Each half-life reduces the amount by half, so after 5 half-lives, it would be reduced by a factor of 2^5 = 32.
The half-life of cesium-137 is approximately 30.1 years, not 2 years. After one half-life, 5 G of the original 10 G sample would remain. After two half-lives (about 60.2 years), 2.5 G would remain, and so on. If you meant a hypothetical isotope with a 2-year half-life, after 2 years, 5 G would remain, and after 4 years, 2.5 G would remain.
The strength of the Union is what caused the three border states to remain in the Union. The states had to take a look at the fighting sides and decide which they felt would benefit them the most at the end of the war.
After 8.1 days, three half-lives have passed (8.1 days / 2.7 days = 3). With each half-life, the number of atoms is halved. Therefore, starting with 800 atoms, after three half-lives there would be 800 / 2 / 2 / 2 = 100 atoms remaining.
After 6 half lives, the remaining will be (1/2)6 i.e 1/64 th of the initial amount. Hence by percentage it would be 1.5625 %