Question

determine:
a) velocidade
b) o aumento da energia cinética do próton.

Answer 1

a)

$v^2 = v_ {0}^{2} + 2a\Delta s \\ \\ v^2 = {(2,4 \cdot {10}^{7} )}^{2} + 2\cdot3,6\cdot {10}^{15} \cdot0,035 \\ \\ {v}^{2} = 5,76\cdot {10}^{14} + 2,52\cdot {10}^{14} \\ \\ {v}^{2} = 8,28\cdot {10}^{14} \\ \\ v = 2,9\cdot {10}^{7} m/s$

b)

$\Delta E_c = \frac{m( {v}^{2} - v_ {0}^{2} )}{2} \\ \\ \Delta E_c = \frac{1,67 \cdot {10}^{ - 27}( ({2,9\cdot {10}^{7}) }^{2} - {(2,4\cdot {10}^{7} )}^{2} }{2} \\ \\ \Delta E_c = 2,1 \cdot {10}^{ - 13} J$

Answer 2

Explicação passo-a-passo:

a)

Pela equação de Torricelli:

$\sf v^2=(v_0)^2+2\cdot a\cdot\Delta S$

Temos:

• $\sf v_0=2,4\cdot10^{7}~m/s$

• $\sf a=3,6\cdot10^{15}~m/s^2$

• $\sf \Delta S=3,5~cm=3,5\cdot10^{-2}~m$

Assim:

$\sf v^2=(v_0)^2+2\cdot a\cdot\Delta S$

$\sf v^2=(2,4\cdot10^7)^2+2\cdot3,6\cdot10^{15}\cdot3,5\cdot10^{-2}$

$\sf v^2=5,76\cdot10^{14}+25,2\cdot10^{13}$

$\sf v^2=5,76\cdot10^{14}+2,52\cdot10^{14}$

$\sf v^2=(5,76+2,52)\cdot10^{14}$

$\sf v^2=8,28\cdot10^{14}$

$\sf v=\sqrt{8,28\cdot10^{14}}$

$\sf \red{v=2,877\cdot10^7~m/s}$

b)

=> Energia cinética inicial

$\sf E_{c_{1}}=\dfrac{m\cdot(v_0)^2}{2}$

$\sf E_{c_{1}}=\dfrac{1,67\cdot10^{-27}\cdot(2,4\cdot10^7)^2}{2}$

$\sf E_{c_{1}}=\dfrac{1,67\cdot10^{-27}\cdot5,76\cdot10^{14}}{2}$

$\sf E_{c_{1}}=\dfrac{9,6192\cdot10^{-13}}{2}$

$\sf E_{c_{1}}=4,8096\cdot10^{-13}~J$

=> Energia cinética final

$\sf E_{c_{2}}=\dfrac{m\cdot v^2}{2}$

$\sf E_{c_{2}}=\dfrac{1,67\cdot10^{-27}\cdot(2,877\cdot10^7)^2}{2}$

$\sf E_{c_{2}}=\dfrac{1,67\cdot10^{-27}\cdot8,277129\cdot10^{14}}{2}$

$\sf E_{c_{2}}=\dfrac{13,8228\cdot10^{-13}}{2}$

$\sf E_{c_{2}}=6,9114\cdot10^{-13}~J$

O aumento da energia cinética do próton foi:

$\sf ~6,9114\cdot10^{-13}-4,8096\cdot10^{-13}=\red{2,1018\cdot10^{-13}~J}$