Question

question 1 of 15 view policies current attempt in progress calculate the kinetic energy (in j) of a proton moving at a speed of 3400 m/s. j etextbook and media save for later attempts: 0

Explanation:

Step1: Recall the formula for kinetic energy

The formula for kinetic energy ($KE$) is $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. The mass of a proton is approximately $m = 1.67\times 10^{-27}\ \text{kg}$, and the velocity $v = 3400\ \text{m/s}$.

Step2: Substitute the values into the formula

First, calculate $v^2$: $v^2=(3400)^2 = 3400\times3400= 11560000\ \text{m}^2/\text{s}^2$.

Then, substitute $m$ and $v^2$ into the kinetic energy formula:
\[

$$\begin{align*} KE&=\frac{1}{2}\times(1.67\times 10^{-27}\ \text{kg})\times(11560000\ \text{m}^2/\text{s}^2)\\ &= 0.5\times1.67\times 10^{-27}\times1.156\times 10^{7}\\ &=(0.5\times1.67\times1.156)\times10^{-27 + 7}\\ &=(0.835\times1.156)\times10^{-20}\\ &\approx0.965\times 10^{-20}\\ &= 9.65\times 10^{-21}\ \text{J} \end{align*}$$

\]

Answer:

$9.65\times 10^{-21}$ (or a more precise value can be calculated, for example, if we use more precise mass of proton $1.6726\times 10^{-27}\ \text{kg}$, the calculation would be $\frac{1}{2}\times1.6726\times 10^{-27}\times(3400)^2\approx\frac{1}{2}\times1.6726\times 10^{-27}\times1.156\times 10^{7}\approx0.8363\times1.156\times 10^{-20}\approx0.9668\times 10^{-20}=9.67\times 10^{-21}\ \text{J}$)