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This will be possible when setting them up in summer with a certain quantity of sag, they have already know that the cables won't be able to sag any further because of the heat. During winter, when the cables contract because of the cold weather, the sag will therefore be reduced, but much tension will not be put on the cables.

4 0

3 years ago

An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and ind

Misha Larkins [42]

Answer:

The unknown element is Capacitor.

Explanation:

The sinusoidal voltage is given as:

v(t) = V Sin (ωt + Ф)

Where:

V = Amplitude of Voltage

ω = 2π / T = > Time period (T)

Ф = phase shift

Considering no horizontal phase shift in the wave form, the equation can be written as:

v(t) = V Sin (ωt)--------(1)

Since, current in the capacitor can be given as:

i(t) = C dv(t)/dt = ωCV Cos (ωt)--------(2)

Now, checking all conditions:

At t=0 :

Equation (1) implies:

v(t) = V Sin [(2π/T)(0)] = V Sin (0)

v(t) = 0

The above finding satisfies the condition in the question. Now checking other conditions.

At t = T/4:

Equation (2) implies:

i(t) = ωCV Cos [(2π/T)(T/4)] = ωCV Cos [(π/2)] = ωCV (0)

i(t) = 0

At t = T/2 :

Equation (2) implies:

i(t) = ωCV Cos [(2π/T)(T/2)] = ωCV Cos [(π)] = ωCV (-1)

i(t) = - ωCV = max amplitude of current in negative direction

All three conditions of voltage and currents of question are satisfied with equations of capacitor hence, the unknown element is capacitor.

4 0

4 years ago

For the data points (1,1),(2,1/2),(3,1/3),(4,1/4), finde the natural cubic spline.

kow [346]

Answer:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

Explanation:

A cubic has the form:

y = ax³ + bx² + cx + d

Given four points, we can write a system of equations:

1 = a + b + c + d

1/2 = 8a + 4b + 2c + d

1/3 = 27a + 9b + 3c + d

1/4 = 64a + 16b + 4c + d

Solving this algebraically would be time-consuming, but we can use matrices to make it easy.

$\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right]\left[\begin{array}{cccc}a\\b\\c\\d\end{array}\right]=\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right]$

First, we find the inverse of the coefficient matrix. This is messy to do by hand, so let's use a calculator:

$\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right] ^{-1} =-\frac{1}{12}\left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]$

Now we multiply by the solution matrix (again using a calculator):

$-\frac{1}{12} \left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right] =\left[\begin{array}{cccc}-1/24\\5/12\\-35/24\\25/12\end{array}\right]$

So the cubic is:

y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12

7 0

3 years ago