Physics, 21.11.2020 08:50 runninglovexoxo

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why didnt people in marie curies time realize radioactivity could be harmfull?

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Answer from: emiliogutierrez

Health

Throughout her work with radium, Marie was unaware of the effects of radioactivity exposure on the body. In her lab, she would keep tubes of radium in her pocket. [3] She began to suspect that radium negatively impacted health when one of her fellow researchers died of a blood disease, and then a few years later her personal assistant died of a blood disease. Even though she suspected that radium exposure was bad for her health, she did very little to monitor her own blood. In 1932, she broke her wrist and the break took much longer to heal than it should have. She then began to notice that her vision was deteriorating and radiation burns on her fingers were becoming more and more painful. Some days she felt too ill to even go to the lab, and finally on July 4, 1934, Marie died from aplastic anemia. [1]

Radium Exposure Treatments

Marie suspected that her health was being negatively impacted by radium exposure, but did nothing about it, most likely because there weren't any effective treatments for radium poisoning yet. At the time, scientists knew that radium was metabolized like calcium. In an attempt to remove it from the system, they manipulated calcium intake. [4] This caused little to no improvements, so parathyroid hormone was added to the treatment. Again, there was some reduction of radium, but not a significant amount. It wasn't until after Marie's death that they realized once radium is in the bones, it is extremely difficult to extract. The lack of therapies for radium exposure may explain why Marie just ignored her symptoms, because she was fully aware of her fate. [4]

© Jenna Gray. The author warrants that the work is the author's own and that Stanford University provided no input other than typesetting and referencing guidelines. The author grants permission to copy, distribute and display this work in unaltered form, with attribution to the author, for noncommercial purposes only. All other rights, including commercial rights, are reserved to the author.

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Answer from: Quest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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