tukeywilliam

Why does Young's Modulus imply that there are atoms on a wire? In other words, in when stretching the wire, why is it implied that there are atoms?

do you mean: [math]\vec{e_1} = \begin{bmatrix}1\\0\\0\end{bmatrix}, \vec{e_2} = \begin{bmatrix}0\\1\\0\end{bmatrix}, \vec{e_3} = \begin{bmatrix}0\\0\\1\end{bmatrix} [/math]?

Thanks. Also for the second one I found one basis to be: [math] a, 2a+3b [/math] because they are linearly independent? is this right? And we can probably say that if the determinant is 0, then the vectors are linearly dependent because inverse is 0.

Lets say an LR circuit has an inductor and a resistor. This mean that is it non-oscillating right?

For the first one I got the dimension to be 3. Is this correct?

How would you find the dimension of the subspace of [math]\mathbb{P}_{3}[/math] spanned by the subset [math]\{t,t-1,t^{2}+1\}[/math]? How would you find a basis and dimension of the given subspace [math]\mathbb{R}^{3}\}[/math] [math]\{[a,a-b,2a+3b]|a,b, \in \mathbb{R}\}[/math] as a subspace of [math]\mathbb{R}^{3}[/math]?

A block is hung on a spring, and the frequency [math] f [/math] of the oscillation of the system is measured. The block, a second identical block, and the spring are carried into space. The 2 blocks are attached to the ends of a spring, and the system is taken out into space on a space walk. The spring is extended, and the system is released to oscillate while floating in space. What is the frequency of oscillation for this system, in terms if [math] f [/math] I know the answer is [math]\sqrt{2}f [/math] . But how do we get this? Thanks

For a vertical spring, if we let [math] x_s [/math] be the total extension of the spring from its equilibrium position without a hanging object, then why does [math] x_s = -\left(\frac{mg}{k}\right) + x [/math]? Is this saying that the extension of the spring is changed for a vertical spring when a weight is added on? Does it follow from: [math] -kx = mg [/math] [math] x = -\left(\frac{mg}{k}\right) [/math]? Why is there a [math] +x [/math] in the [math] x_s [/math] expression? Thanks

But why does the determinant involve taking derivatives?

I know that a set of vector functions [math]{\vec{v_{1}}(t), \vec{v_{2}}(t)+...+\vec{v_{n}}(t)}[/math] in a vector space [math]\mathbb{V}[/math] if [math]c_{i}= 0 [/math] for the following equation: [math]c_{1}\vec{v_{1}}(t)+c_{2}\vec{v_{2}}(t)+...+c_{n}\vec{v_{n}}(t) \equiv \vec{0}[/math] Where does the Wronskian come into play? Is it basically a determinant with functions and derivatives? Thanks