Question

Check my work please?

Answer 1

Problem: What is the speed of a proton that has been accelerated from rest through a potential difference of -1000 V?

Solution: \[U_{e}= K_{e}\]
\[U = \frac{ 1 }{ 2 }m_{p}V^2\]
\[v=\frac{ 2U }{ m_{proton} }\]

Answer 2

Sorry forgot to square root the right side

Answer 3

I get that v = 1.09 m/s ???

Answer 4

@agent0smith

Answer 5

correct

Answer 6

\(\Huge \checkmark\)

Answer 7

Your velocity doesn't seem right?
http://www.wolframalpha.com/input/?i=sqrt(2*1.6*10%5E(-19)*1000%2F(1.67*10%5E(-27)))

Answer 8

mathmale

Answer 9

\[\large v=\sqrt{\frac{ 2U }{ m_{proton} }}=\sqrt{\frac{ 2qV }{ m_{proton} }}\]where q=charge of electron/proton 1.6*10^(-19),
V=1000,
mass of proton 1.67×10^(-27)