Answer:
D) biophysics.
Explanation:
Hello,
In this case, since biophysics is the study of physical processes and phenomena happening in living things at molecular scale, which includes cells, tissues and organisms. Both experimental and theoretical techniques are used in order to understand biological and biochemical behaviors inside and outside the cells.
Best regards.
Answer:
a) 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b) 3.466 × 10¹¹ N/C
Explanation:
a)
p(r) = -A exp ( - 2r/a₀)
Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV = -A ₀∫^∞ ₀∫^π ₀∫^2π exp ( - 2r/a₀)r² sinθdrdθd∅
Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e
now using integration by parts;
A = e / πa₀³
p(r) = - (e / πa₀³) exp (-2r/a₀)
Now Net charge inside a sphere of radius a₀ i.e Qnet is;
= e - (e / πa₀³) ₀∫^a₀ ₀∫^π ₀∫^2π r² exp (-2r/a₀)dr
= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b)
Using Gauss's law,
E × 4πa₀ ² = Qnet / ∈₀
E = 4πa₀ ² × Qnet × 1/a₀²
E = 3.466 × 10¹¹ N/C
Answer:
Explanation:
F = kQq/r²
r = √(kQq/F)
a) r = √(8.899(10⁹)(8)(4) / 18(10¹³)) = 0.0397749... m
r = 40 mm
b) r = √(8.899(10⁹)(12)(3) / 18(10¹³)) = 0.0421876... m
r = 42 mm
Answer:
27.14 ohms
Explanation:
V = IR
V/I = R
1.9 v / .07 a = 27.14 ohms