Answer:
a) q_1=q_2= 7.42*10^-7 C
b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C
Explanation:
Given:
F_e = 0.220 N
separation between spheres r = 0.15 m
Electrostatic constant k = 8.99*10^9
Find: charge on each sphere
part a
q_1 = q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_1^2 = F_e*r^2/k
q_1=q_2= sqrt (F_e*r^2/k)
Plug in the values and evaluate:
q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)
q_1=q_2= 7.42*10^-7 C
part b
q_1 = 4*q_2
Using coulomb's law:
F_e = k*q_1*q_2 / r^2
q_2^2 = F_e*r^2/4*k
q_2= sqrt (F_e*r^2/4*k)
Plug in the values and evaluate:
q_2= sqrt (0.22*0.15^2/4*8.99*10^9)
q_2= 3.7102*10^-7 C
q_1 = 14.8*10^-7 C
Answer:
12 m/s
Explanation:
Using the continuity equation, which is an extension of the conservation of mass law
ρ₁A₁v₁ = ρ₂A₂v₂
where 1 and 2 indicate the conditions at two different points of flow, in this case, point 1 is any normal position in the pip and point 2 is the conditions at the restriction.
ρ = density of the fluid flowing; note that the density of the fluid flowing (water) is constant all through the fluid's flow
A₁ = Cross sectional Area of the pipe at point 1 = (πD₁²/4)
A₂ = Cross sectional Area of the pipe at the restriction = (πD₂²/4)
v₁ = velocity of the fluid flowing at point 1 = 3 m/s
v₂ = velocity of the fluid flowing at The restriction = ?
ρ₁A₁v₁ = ρ₂A₂v₂
Becomes
A₁v₁ = A₂v₂ (since ρ₁ = ρ₂)
(πD₁²/4) × 3 = (πD₂²/4) × v₂
3D₁² = D₂² × v₂
But
D₂ = (D₁/2)
And D₂² = (D₁²/4)
3D₁² = D₂² × v₂
3D₁² = (D₁²/4) × v₂
(D₁²/4) × v₂ = 3D₁²
v₂ = 4×3 = 12 m/s
(31-15 = 16).
Explanation:
the element phosphorus (P) has an atomic number of 15 and a mass number of 31. Therefore, an atom of phosphorus has 15 protons, 15 electrons, and 16 neutrons
Well, in some systems the atoms melt or burn or change state like from liquid to gases when they reach a certain temperature. This could decrease the energy in the system.
Hope I helped
