Answer:
Explanation:
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By firing electrons from a cathode gun at a known velocity through an electric field of a known strength, J.J Thomson was able to observe how much the field deflected the beam. With a mass-to-charge ratio calculated, he could determine that the mass of the electron was not significantly heavy (in fact almost nothing compared to the proton or neutron)
The biggest thing you're doing wrong is ignoring the units
when you're working with the quantities.
Now let's look at the rest of the problem:
The formula you used is correct:
Net flux through the surface = (net charge inside) / ε₀
and ε₀ = 8.85 x 10⁻¹² farad/meter.
What's the net charge inside the surface in this problem ?
It's (5.85 x 10⁷ electrons) x (the charge on each electron)
= (5.85 x 10⁷ electrons) x (-1.6 x 10⁻¹⁹ coulomb/electron)
= -9.36 x 10⁻¹² coulomb .
Finally, (net charge inside) / ε₀
= (-9.36 x 10⁻¹² coulomb) / (8.85 x 10⁻¹² farad/meter)
= -1.058 newton-m²/coulomb .
The sign and the significant figures in your answer are correct, so
we can see that you know what you're doing. The only thing left is
the order of magnitude. You most likely took one of the negative
exponents and made it positive. You got an answer that's 10²² too
small. Big deal. You could claim "that's close", and see whether you
can convince a teacher.
Answer:
n=142 N
Explanation:
solution:
pic 1 is attached
<em>There is no vertical acceleration (ay = 0), so net sum of vertical forces is 0. n acts up, weight and the vertical component of F act down, so </em>
∑F_y =m*a_y
n-mg-Fsin43°=0
n=142 N