6 Inductance‚ Capacitance‚ and Mutual Inductance Assessment Problems AP 6.1 [a] ig = 8e−300t − 8e−1200t A v=L dig = −9.6e−300t + 38.4e−1200t V‚ dt 38.4e−1200t = 9.6e−300t t > 0+ v(0+ ) = −9.6 + 38.4 = 28.8 V [b] v = 0 when or t = (ln 4)/900 = 1.54 ms [c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W dp = 0 when e1800t − 12.5e900t + 16 = 0 [d] dt Let x = e900t x = 1.45‚ x = 11.05‚ and solve the quadratic x2 − 12.5x + 16 = 0 t= ln 1.45 = 411.05 µs 900 ln 11.05 = 2.67 ms 900 t= p is
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triangle ABC‚ sides AB‚ BC‚ and CA are represented by the small letters c‚ a and b. These are used as it is a rule to denote a side opposite to a particular angle in a triangle with the same letter in lower case. Thus the formula is given by – a2 = b2 + c2 – 2bc(CosA) Similarly‚ other sides can also be found
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Physics 103 Equations a2 + b2 = c2 If v = ∆x ¯ ∆t |g| = 9.8m/s2 1 K = 2 mv 2 Wnet = ∆K 2 |acent | = vtang /r ω = 2πf = 2π T sin θ = opposite hypotenuse cos θ = adjacent hypotenuse tan θ = sin θ/ cos θ b x = −b± 2a −4ac v = v0 + at p = mv ¯ P =W t Wcons = −∆U P I = 4πr2 Fspr = −k ∆x √ 2 ax2 + bx + c = 0 then 1 a = ∆v ¯ ∆t x = x0 + v0 t + 2 at2 Fext = ma = ∆p w = mg ∆t W =F s Ug = mgy ME = U + K M Ei + Wother = M Ef |τ | = rF⊥ v = fλ P = F/A Uspr = 1 kx2 2 1 4π 0 e = 1.6 × 10−19
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r is the distance between the centre O and any point (such as A) of circle C1. Figure 1 The second diagram shows circle C2 with radius OP and centre P‚ as well as circle C3 with radius r and centre A. An intersection between C1 and C2 is marked by point A. The intersection of C3 with OP is marked by point P’. Figure 2 Through this investigation I will examine how the r values correlate with the values of OP in determining
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arr[i]=arr[pos_min]; c8 0 arr[pos_min]=temp; c9 0 c1n + c2(n-1) + c3n^2 + c4(n^2-1) + c5(0) + c6(n^2-1) + c7(0) + c8(0) + c9(0) = (c3 + c4 + c6)n^2 + (c1 + c2)n - (c2 + c4 + c6) This is quadratic an^2 + bn + c --------------------------- Worst Case Scenario cost times for(int i=0; i< arr[pos_min]) c4 n^2-1 pos_min = j; c5 n^2-1 if(pos_min != i) c6 n^2-1 temp=arr[i]; c7 n^2-1 arr[i]=arr[pos_min]; c8 n^2-1 arr[pos_min]=temp; c9 n^2-1 c1n + c2(n-1) + c3n^2 + c4(n^2-1) + c5(n^2-1) + c6(n^2-1)
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Exploring Rates‚ Equilibrium‚ and Organic Chemistry Brandon Sher SCH 4U For: Ms. Merhai Due date: June 1st‚2013 Introduction: Organic chemistry. Two quantities one could measure with the presence of organic chemicals are chemical equilibrium and rate of reaction. Chemical equilibrium is the ratio between the forward and reverse process of a reaction. This is represented by the equation: K = products / reactants. When a reaction has reached equilibrium‚ the
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that is God. An example of this argument is the Pros logion of St. Anselm where the monk tried to provide Christian documentation for the existence of god. The relationship between C1 and C2 can be identified as a part-whole relationship where C2 can be considered part of C1. The reason being is because C2 refers to this idea of god and how people using different doctrines try to prove this existence but in order to prove his existence you have to include this idea of rationalism because you need
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Because of the policies of the government‚ the currency of Venezuela‚ the Bolivar has and continues to devalue to a point that it is affecting US companies already doing business in Venezuela. PepsiCo and CocaCola took $126 and $660 million loses respectively‚ as a result of the devaluation of the Bolivar (Gillespie Feb 11th 2015). Of the 178 countries ranked by the Heritage Index of Economic Freedom‚ Venezuela ranks 176th (Venezuela 2015). Venezuela may
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that a system of algebraic equations can be expressed in the form of matrices. This means‚ a system of linear equations like a1 x + b1 y = c 1 a2 x + b2 y = c 2 ⎡ a b ⎤ ⎡ x ⎤ ⎡c ⎤ can be represented as ⎢ 1 1 ⎥ ⎢ ⎥ = ⎢ 1 ⎥ . Now‚ this ⎣ a2 b2 ⎦ ⎣ y ⎦ ⎣ c2 ⎦ system of equations has a unique solution or not‚ is determined by the number a1 b2 – a2 b1. (Recall that if a1 b1 or‚ a1 b2 – a2 b1 ≠ 0‚ then the system of linear ≠ a2 b2 equations has a unique solution). The number a1 b2 – a2 b1 P.S. Laplace
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associated with c1=4‚ c2=4. From this‚ one can obtain the sample size n=28.6≅29‚ thus‚ the selected plan parameters of RGRSS is n=29‚ c1=4and c2=4. Example 2‚ for given p2=0.002 and h2=0.002 from Table-1. Column headed h2‚ locate the value which is equal to or just greater than the specified h2 which is 0.050071‚ corresponding to this h2 and np2. Which associated with c1=1‚ c2=2. From this‚ one can obtain the sample size n=25‚ thus‚ the selected plan parameters of RGRSS is n=25‚ c1=1 and c2=1. Example 3
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