Question

Calculate the speed of a proton after it accelerates from rest through a potential difference of 380 v .

Answer 1

The magnitude of the change in potential energy is equal to the kinetic energy,

$\left | \Delta U \right |= K$

or                        

$qV=\frac{1}{2} mv^{2}$    

Here, V is potential difference, q is charge, m is mass and v is velocity.

We can also write,

$v=\sqrt{\frac{2qV}{m} }$

Given  $V=380 V$

Substituting this value with mass of proton, $m=1.672\times10^{-27} kg$ and charge of proton,$q= 1.6\times10^{-19} C$ we get

$v=\sqrt{\frac{2\times1.6\times10^{-19}C \times 380V}{1.672\times10^{-27} kg} }\\\\v= 727.27\times10^{8} m/s =7.27\times10^{6}m/s$

Therefore, the speed of proton is $7.27\times10^{6}m/s$.

Answer 2

Answer:

The speed of a proton after it accelerates from rest is 2.698x10⁵ m/s

Explanation:

Please look at the solution in the attached Word file

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