The question seems to involve a misunderstanding or is improperly formed for a couple of reasons:

1. When you refer to a right-angled triangle with “sides of 90 degree unit,” it suggests a confusion. In geometry, the sides of a triangle are measured in units of length (not degrees, which measure angles). A right-angled triangle is defined by having one angle measuring 90 degrees, but the lengths of the sides are not described in degrees.

2. The functions L = cos y and M = sin x appear to introduce variables y and x as angles, but without specific values or a clear connection to the triangle’s sides, they cannot directly contribute to finding the area of the triangle. Normally, to find the area of a right-angled triangle, you need the lengths of two sides that meet at the right angle (often referred to as the base and the height), and then you use the formula:

[ text{Area} = frac{1}{2} times text{base} times text{height} ]

Without specifying the lengths of the triangle’s sides or how the functions L and M relate to those lengths (for instance, if they represent the triangle’s angles or if they somehow define the lengths of sides in relation to an angle), it’s not possible to provide an answer that integrates all given information directly.

If there’s a specific right-angled triangle scenario with known side lengths or specific angles (apart from

To find the resistivity ((rho)) of a material, we can use the formula:

[ rho = R times frac{A}{L} ]

where (R) is the resistance, (A) is the cross-sectional area, and (L) is the length of the material.

Given that the resistance ((R)) is (200 , Omega), the length ((L)) is (10 , m), and the area ((A)) is twice that of the length, there seems to be a misunderstanding in how the area is described. The area cannot be directly twice the length as they are of different dimensions. Instead, if the intended meaning is that the area is related to the dimensions of the length in some specific manner that is not clearly described, we’ll need a clearer understanding to proceed accurately. For instance, if the area is implied to be a function of a dimension that can be related back to the length, we would need that specific relation described (e.g., if it’s twice the cross-sectional dimension related to the length, we still need to know the shape or further details to calculate it).

However, to proceed with an attempt to interpret your request, we’ll assume a simplistic approach where perhaps what was meant is that the cross-sectional area is somehow numerically ‘twice’ in some unit of measure without direct correlation to meters since the dimensional units must match appropriately for such calculations. Since this

To find the potential (V) at (x = 0.5 , text{mm}), given that the potential (V) at (x = 1 , text{mm}) is (2 , text{V}), and the volume charge density (rho_v) is (-10^6 varepsilon_0), constant throughout the free space region between (x = 0) and (x = 1 , text{mm}), we’ll use the concept of electric field and potential in a region with constant volume charge density.

Given data:

– (rho_v = -10^6 varepsilon_0) (Volume charge density, where (varepsilon_0) is the permittivity of free space (left(varepsilon_0 = 8.85 times 10^{-12} , text{F/m}right))).

– (V(x = 1 , text{mm}) = 2 , text{V})

– (V(x = 0) = 0 , text{V}) (Since it’s mentioned that the potential is zero at (x = 0).)

– (x = 0.5 , text{mm}) is the point where we need to find the potential.

To find the potential at (x = 0.

To compute the charge enclosed by the cube using the given electric flux density ( mathbf{D} = frac{10y^3}{3} hat{mathbf{j}} ) C/m(^2), we need to apply Gauss’s Law in integral form, which relates the electric flux through a closed surface to the charge enclosed by that surface. However, because we’re dealing with the electric flux density (mathbf{D}) directly, we can integrate (mathbf{D}) over the surface of the cube to find the total charge enclosed without explicitly invoking Gauss’s Law.

Given that the cube has its edges parallel to the axes and is centered at the origin with edge length 2 m, it extends from (-1) m to (1) m along the x, y, and z axes.

Since (mathbf{D}) only has a y-component ((frac{10y^3}{3} hat{mathbf{j}})), the flux through the cube will only occur through the faces perpendicular to the y-axis, i.e., the faces at (-1) m and (1) m in the y-direction. The areas of the faces through which (mathbf{D}) passes are parallel to the xz-plane.

The total charge enclosed ((Q_{text{enc}})) by the cube can be obtained by integrating the normal component of (math

To compute the divergence of a vector field using the divergence theorem in spherical coordinates, we follow a systematic approach. Given a vector field (mathbf{D} = frac{5r^2}{4} hat{i}) in a coordinate system, we first need to express this field in spherical coordinates and then find its divergence. However, a direct conversion of the given vector field into spherical coordinates poses a challenge since the field is given in a form that suggests it’s already partially in a non-Cartesian form ((frac{5r^2}{4} hat{i}) suggests a dependence on radial distance but uses (hat{i}), which is a Cartesian unit vector). Assuming the intention is to deal with a radially dependent vector field in a spherical context, we can reinterpret the vector field in spherical coordinates, focusing on its radial component only.

Spherical Coordinates Background

In spherical coordinates, a position in space is given by (r) (radial distance), (theta) (polar angle, measured from the positive z-axis), and (phi) (azimuthal angle, measured in the x-y plane from the positive x-axis). Vector fields in spherical coordinates are expressed in terms of these variables and their unit vectors (hat{r}), (hat{theta}), and (hat{phi}).

Given Vector Field

Given the ambiguities in the initial presentation of the vector field, but