v₀ =5•10⁶ m/s
v=2.1•10⁶ m/s
s=2•10⁻⁴ = 2•10⁻⁶ m
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v= v₀-at
a= (v₀-v)/t
substitute 'a' in
s=v₀t+at²/2
and solve for 't'
An electron moving at a speed of 5 x 10 raise to power 6m/s was shot through a sheet of paper which is 2.1 x 10 raise to power 4cm thick. The electron emerges from the paper with a speed of 2 x 10 raise to power 6 m/s. Find the time taken by the electron to pass through the sheet.
Pls show full working and formular.
5 answers
substitute 'a' in
s=v₀t - at²/2
and solve for 't'
s=v₀t - at²/2
and solve for 't'
Please save yourself the extra time of writing scientific notation numbers such as
5*10^6
Instead of
5 x 10 raise to power 6.
It is also easier to read.
In the case of this problem, divide the thickness of the paper by the AVERAGE speed of the electron as it passes through. That average speed is
3.5*10^6 m/s.
I think you have omitted an important minus sign. I have never seen a sheet of paper 210 meters thick.
5*10^6
Instead of
5 x 10 raise to power 6.
It is also easier to read.
In the case of this problem, divide the thickness of the paper by the AVERAGE speed of the electron as it passes through. That average speed is
3.5*10^6 m/s.
I think you have omitted an important minus sign. I have never seen a sheet of paper 210 meters thick.
thanks
v=5×10^6
u=2×10^6
s=2.1×10^-6
v^2-u^2=2as==>(5×10^6)^2-(2×10^6)^2=2×a×2.1×10^-6
a=5.2×10^18
v=u+at==>5×10^6=2×10^6+5.2×10^18×t
t=0.6×10^-12
t=6×10^-13
u=2×10^6
s=2.1×10^-6
v^2-u^2=2as==>(5×10^6)^2-(2×10^6)^2=2×a×2.1×10^-6
a=5.2×10^18
v=u+at==>5×10^6=2×10^6+5.2×10^18×t
t=0.6×10^-12
t=6×10^-13
1 answer