A water wave is an example of a mechanical wave. A wave that can travel only through matter is called a mechanical wave.
<h3><u>Answer;</u></h3>
= 73 N
<h3><u>Explanation</u>;</h3>
Using the formula
2 T cos(30°) = w
Where; T is the tension on each string, while w is the weight of the box given by mg
Therefore;
W = 2Tcos 30°
= 2 × 42 cos 30°
= 84 cos 30°
= 72.74
<u> ≈ 73 N</u>
The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is
<span>dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) </span>
<span>where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. </span>
<span>However, cos(T) = x/sqrt(x^2 + R^2) </span>
<span>so the equation becomes </span>
<span>dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) </span>
<span>dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 </span>
<span>Integrating around the ring you get </span>
<span>E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 </span>
<span>E = (R/2*e0)*x*(x^2 + R^2)^-1.5 </span>
<span>we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} </span>
<span>to find the maxima set this = 0, giving </span>
<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>
<span>mult both side by (x^2 + R^2)^2.5 to get </span>
<span>(x^2 + R^2) - 3*x^2 = 0 </span>
<span>-2*x^2 + R^2 = 0 </span>
<span>-2*x^2 = -R^2 </span>
<span>x = (+/-)R/sqrt(2) </span>
Idk since it looks sooooo close ,but is it is that’s really cool
