Physics, 26.03.2021 21:30 katlyn10289

Si una bala de 122 g que viaja a 350 m/s golpea a un policía con chaleco antibalas el cual soporta 120 J de energía, ¿la bala atraviesa o no el chaleco? Explicar brevemente.

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Puesto que la energia cinética traslacional es mucho mayor que la capacidad del chaleco antibalas, la bala atravesaría el chaleco antibalas.

Explanation:

Un chaleco antibalas soporta el disparo de una bala disipando la energía de esta última a través de su propio material. Si sabemos que el chaleco antibalas soporta 120 joules de energía, cabe saber si la energía cinética traslacional es igual o inferior a ese límite, significando que la bala no atravesaría el chaleco.

La energía cinética traslacional de la bala (), in joules, queda expresada con la siguiente fórmula:

$K = \frac{1}{2}\cdot m\cdot v^{2}$ (1)

Donde:

- Masa de la bala, en kilogramos.

- Rapidez de la bala, en metros por segundo.

Si sabemos que $m = 0.122\,kg$ y $v = 350\,\frac{m}{s}$ , entonces la energía cinética traslacional de la bala es:

$K = 7472.5\,J$

Puesto que la energia cinética traslacional es mucho mayor que la capacidad del chaleco antibalas, la bala atravesaría el chaleco antibalas.

Answer from: Quest

a.) is the correct answer.

explanation:

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

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