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Allisa [31]
3 years ago
13

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia

lly uncharged) with plates 5.0 mm apart. The switch is closed at t =0s .
Required:
a. After the switch is closed, find the maximum electric flux through the capacitor.
b. After the switch is closed, find the maximum displacement current through the capacitor.
c. Find the electric flux at t =0.50ns.
d. Find the displacement current at t =0.50ns.
Physics
1 answer:
slega [8]3 years ago
8 0

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

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Calculate the molarity of a 10. 0% cacl₂ solution. The density of the solution is 1. 0835 g/cm³.
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The molarity of 10% CaCl2 is 0.9%

concentration of the given salt CaCl₂ = 10%

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1 year ago
A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t
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Answer:

a)    F₁ = 267.3 N,   N₁ = 1300 N,  b)    μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

           

let's use subscript 1 for the ladder and 2 for the firefighter

            ∑ τ = 0

          -W₁ x₁ - W₂ x₂ + N₁ y = 0

           N₁ = \frac{W_1 x_1 + W_2 x_2}{y}          (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

         cos 60 = x₁ / d₁

         x₁ = d₁ cos 60

         x₁ = 7.5 cos 60

         x₁ = 3.75 m

for the firefighter d₂ = 4 m

         cos 60 = x₂ / d₂

         x₂ = d₂ cos 60

          x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

         sin 60 = y / d₃

         y = d₃ sin 60

         y = 15 sin 60

         y = 13 m

we look for the Normal by substituting in equation 1

         N₂ = \frac{500 \ 3.75 \ + 800 \ 2}{13}

         N₂ = 267.3 N

now let's use the translational equilibrium relations

 X axis

           F₁ - N₂ = 0

           F₁ = N₂

           F₁ = 267.3 N

Axis y

          N₁ - W₁ -W₂ = 0

          N₁ = W₁ + W₂

          N₁ = 500 + 800

          N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

          x₂ = 9 cos 60

          x₂ = 4.5 m

we substitute in 1

          N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}  

          N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

          fr = F₁ = N₂

          fr = 421.15 N

friction force has the expression

          fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

          μ = fr / N₁

          μ = 421.15 / 1300

          μ = 0.324

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