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:
Explanation:
Range= U²sin2theta/g
= 102²* sin (2*42)/9.8
= 1054.74m
Yes she will be abable to jump
The distance from the edge will be
(1054.74- 420)m
= 634.74m
Answer:
a)
85.05 N/m
b)
179.81 rad/s
Explanation:
a)
k = spring constant of the spring
m = mass of the block = 0.473 kg
x = stretch caused in the spring = 0.109 m
h = height dropped by the block = 0.109 m
Using conservation of energy
Spring potential energy gained by the spring = Potential energy lost by the block
(0.5) k x² = mgh
(0.5) k x² = mgx
(0.5) k x = mg
(0.5) k (0.109) = (0.473) (9.8)
k = 85.05 N/m
b)
angular frequency is given as
= 179.81 rad/s
You know that the formula for finding horizontal displacement is;
s=uₓt
s=(5 m/s)(3s)
s=15 m
Therefore, the ball travelled 15 m.
Hope I helped :)