Physics, 05.09.2020 01:01 rsanchez1226

Which form of emission is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass numbers?

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Answer from: gfdsgfd5654

Gamma radiation .

Explanation:

Three types of emission result at the time of nuclear emission .

alpha emission

Beta emission

Gamma radiation emission

Gamma radiation results in the emission of gamma ray . This is a very high frequency radiation . It is not a particle having charge and mass like alpha and beta emission . So it has no effect on the mass and charge of the product that is formed after gamma radiation . The only effect it has on the product is that it reduces the kinetic energy of the emitting particles or it reduces the internal energy of the product . It causes negligible change in the mass of the product and no change in the charge of the product.

That is why , generally this product is not written in the nuclear reaction .

Answer from: Quest

Ithink the answer is d i hope it my friend message me if i’m wrong and i’ll change my answer and fix it for you

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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