Answer:
995.12 N/C
Explanation:
R = 9 cm = 0.09 m
σ = 9 nC/m^2 = 9 x 10^-9 C/m^2
r = 9.1 cm = 0.091 m
q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C
E = kq / r^2
E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)
E = 995.12 N/C
The recessive trait will always show up
<span>The waves with the lowest energy and lowest frequencies of the electromagnetic spectrum are the "Radio waves"
So, option B is your answer
Hope this helps!
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I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>
