answer:Given: - Mass of object A ({eq}m_A {/eq}): 1.5 kg - Mass of object B ({eq}m_B {/eq}): 3.4 kg - Height of the pulley ({eq}h_1 {/eq}): 4.8 m - Initial height of both masses ({eq}h_2 {/eq}): 1.8 m - Acceleration due to gravity ({eq}g {/eq}): 9.8 m/s² Since the pulley is massless and the cord is frictionless, the tension in the cord is the same for both masses, and they have the same acceleration ({eq}a {/eq}). Applying Newton's second law for both masses, we get: For mass A: {eq}T - m_A g = m_A a {/eq} For mass B: {eq}m_B g - T = m_B a {/eq} Combining the equations, we find the acceleration: {eq}a = frac{m_B - m_A}{m_B + m_A} g {/eq} {eq}a = frac{3.4 , kg - 1.5 , kg}{3.4 , kg + 1.5 , kg} times 9.8 , m/s^2 {/eq} {eq}a = 3.8 , m/s^2 {/eq} The velocity of the heavier mass (B) when it hits the ground is: {eq}v_f^2 = 2 a h_2 {/eq} Since both masses start from rest, the lighter mass (A) will also have the same final velocity. To find its maximum height, we consider the upward motion after it has passed the initial height ({eq}h_2 {/eq}). The maximum height ({eq}Delta x_{max} {/eq}) is twice the initial height: {eq}Delta x_{max} = 2h_2 {/eq} {eq}Delta x_{max} = 2 times 1.8 , m {/eq} {eq}Delta x_{max} = 3.6 , m {/eq} The lighter object reaches a maximum height of 3.6 meters.

answer:The torque produced by the magnetic field on the conducting wire is given by: {eq}vec{tau}=vec{mu}times vec{B} {/eq}, where {eq}vec{mu} {/eq} is the magnetic moment and {eq}vec{B} {/eq} is the magnetic field. The magnetic moment is given by: {eq}vec{mu}=i vec{A} {/eq}, where {eq}vec{A} {/eq} is the area vector perpendicular to the plane of the square. Since the current flows from A to B to C to D, the area vector is in the positive z-direction: {eq}vec{A}=(1 hat k) ;rm m^2 {/eq}. Therefore, the magnetic moment is: {eq}vec{mu}=(0.5 hat k) ;rm Acdot m^2 {/eq}. The torque is then: {eq}begin{align} vec{tau}&=vec{mu}times vec{B}&=(0.5 hat k) ;rm Acdot m^2 times (0.1 hat i + 0.2 hat j - 0.3 hat k) ;rm T &=begin{vmatrix} hat i & hat j & hat k 0 & 0 & 0.5 0.1 & 0.2 & -0.3 end{vmatrix} &=boxed{(-0.1 hat i + 0.05 hat j) ;rm Ncdot m}. end{align} {/eq}

answer:Row operations are performed on a matrix primarily to transform it into a simpler form, such as reduced row echelon form (RREF), which aids in solving systems of linear equations. When we perform row operations, we do not change the row space of the matrix, but the column space can be affected. This is because row operations can be seen as transformations in the space spanned by the column vectors. 1) Row operations do not change the row space because they preserve linear combinations of the rows. However, they can change the column space because the linear combinations of the columns, which define the column space, are affected by the transformation. 2) We typically perform row operations instead of column operations because vectors are conventionally represented as rows in a matrix, making it more intuitive to work with row operations when solving systems of linear equations. If needed, you can take the transpose of the matrix and perform column operations, which is equivalent to row operations on the original matrix. 3) The linear independence of the rows may change due to row operations, but the row space remains the same as long as a new set of linearly independent rows can be found to span the same subspace. Conversely, the linear independence of the columns is preserved because row operations effectively change the representation of the column vectors but not the linear relationships among them. 4) The row space and column space are interconnected: every matrix transforms its row space onto its column space. Geometrically, the row space represents the set of all possible linear combinations of the matrix's row vectors, which can be visualized as the set of all vectors that the matrix maps to non-zero vectors. In essence, the row space represents the directions in which the matrix has non-zero effects. In summary, row operations primarily affect the column space and the linear independence of the rows, but they do not change the row space itself. The connection between the row space and column space lies in their relationship under matrix transformations, and the row space can be geometrically interpreted as the set of vectors that the matrix projects non-trivially.

answer:The Jacobian matrix of the vector-valued function mathbf{r}(x, y, z) is given by: text{Jacobian} = frac{partial mathbf{r}}{partial (x, y, z)} = begin{bmatrix} frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z} frac{partial g}{partial x} & frac{partial g}{partial y} & frac{partial g}{partial z} frac{partial h}{partial x} & frac{partial h}{partial y} & frac{partial h}{partial z} end{bmatrix} = begin{bmatrix} frac{1}{x} & 0 & 0 0 & frac{1}{2sqrt{y}} & 0 0 & 0 & -sin(z) end{bmatrix}