Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Answer:
over a rive or fast moving water or canyon
Explanation: you would use a suspension bridge in an area where you can't put supports down.
Answer:
0.023 Pa*s
Explanation:
The surface area of the side of the inner cylinder is:
A = π*d*l
A = π*0.15*0.75 = 0.35 m^2
At 200 rpm the inner cylinder has a tangential speed of:
u = w * r
u = w * d/2
w = 200 rpm * 2π / 60 = 20.9 rad/s
u = 20.9 * 0.15 / 2 = 1.57 m/s
The torque is of 0.8 N*m, this means that the force is:
T = F * r
F = T / r
F = 2*T / d
For Newtoninan fluids with two plates moving respect of each other with a fluid between the viscous friction force would be:
F = μ*A*u / y
Where
μ: viscocity
y: separation between pates
A: surface area of the plates
Then:
2*T / d = μ*A*u/y
Rearranging:
μ = 2*T*y / (d*A*u)
μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s
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