 # Review. An electron of mass 9.11 \times 10−31 kg has an initial speed of 3 oliviayychengwh 2021-12-20 Answered
Review. An electron of mass $9.11×10-31$ kg has an initial speed of 3.00 $3×105\frac{m}{s}$. It travels in a straight line, and its speed increases to $7.00×105\frac{m}{s}$ in a distance of 5.00 cm. Assuming its acceleration is constant, (a) determine the magnitude of the force exerted on the electron and (b) compare this force with the weight of the electron, which we ignored.
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Ste 1
Given data:
$m=9.11×{10}^{-31}kg$
${v}_{i}=3\mid ×{10}^{5}\frac{m}{s}$
${v}_{f}=7×{10}^{5}\frac{m}{s}$
$\mathrm{△}x=5cm$

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Step 2
(a)
First: we determine the acceleratoin by using the following equation:
${v}_{\left\{f\right\}}^{2}={v}_{\left\{i\right\}}^{2}+2a\mathrm{△}x$
Solving for (a) $a=\frac{{x}_{\left\{f\right\}}^{2}-{v}_{\left\{i\right\}}^{2}}{2\mathrm{△}x}=\frac{\left(7×{10}^{5}\right)-{\left(3×{10}^{5}\right)}^{2}}{2×0.05}=4×{10}^{12}\frac{m}{{s}^{2}}$
Then: we use Newtons
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Step 3
(b)
The weight of the electron is determined by:
$Fg=mg=\left(9.11×{10}^{-31}\right)\left(9.8\right)=8.93×{10}^{-30}N$
Therefore, this force is larger that the weight of the electron by $\frac{3.64×{10}^{-18}}{8.93×{10}^{-30}}=4.08×{10}^{11}$ times.
So,
(a) $\sum F=3.64×{10}^{-18}N$
(b)This force is larger that the weight of the electron by $4.08×{10}^{11}$ times.