Momentum of a body is calculated by multiplying the mass of a moving body with its velocity. When a body is at rest it has zero momentum since the velocity is also zero.
In this case the momentum of the canoe will be;
38 kg × 2.2 m/s = 83.6 kgm/s
Therefore, the correct answer is 83.6 kg m/s
Using the kinematic equation,
v = u + at
26 = 11 + 3t
15 = 3t
t = 5 seconds.
Time required is 5 seconds.
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>
Mass and velocity I think
Voltage=Energy/Charge, V=E/Q
V in volts V, E in joules J, Q in coulombs C.
or
Voltage= Current×Resistance, V=IR
V in volts V, I in amperes A and R in ohms Ω
