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The average distance an electron travels between collisions is 2.0 μm . What acceleration must an electron have to gain 2.0×10−18 J of kinetic energy in this distance? Express your answer to two significant figures and include the appropriate units. a = nothing nothing

Answer :

opudodennis

Answer:

[tex]1.1\times 10^{18} m/s^{2}[/tex]

Explanation:

Work=Kinetic energy

Fd=KE but we know that F=ma hence

ma d= KE

[tex]a=\frac {KE}{md}[/tex]

Here a is acceleration, d is distance, KE is kinetic energy, m is mass

Taking m as [tex]9.11\times 10^{-31} Kg[/tex], KE as [tex]2\times 10^{-18}[/tex] then d as [tex]2\times 10^{-6}[/tex] then

[tex]a=\frac {2\times 10^{-18}}{ 9.11\times 10^{-31}\times 2\times 10^{-6}}=1.1\times 10^{18} m/s^{2}[/tex]

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