Center.........................
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
Answer:
I = 2.667 kg m²
Explanation:
The moment of inertia of a body can be calculated by the expression
I = ∫ L² dm
For high symmetry bodies the expressions of the moment of inertia are tabulated, for a rod with its axis of rotation at its midpoint it is
I = 
m L²
let's calculate
I = 2 4²
I = 2.667 kg m²
Answer:
392 N
Explanation:
Draw a free body diagram of the rod. There are four forces acting on the rod:
At the wall, you have horizontal and vertical reaction forces, Rx and Ry.
At the other end of the rod (point X), you have the weight of the sign pointing down, mg.
Also at point X, you have the tension in the wire, T, pulling at an angle θ from the -x axis.
Sum of the moments at the wall:
∑τ = Iα
(T sin θ) L − (mg) L = 0
T sin θ − mg = 0
T = mg / sin θ
Given m = 20 kg and θ = 30.0°:
T = (20 kg) (9.8 m/s²) / (sin 30.0°)
T = 392 N
Answer:
Magnetic force, F = 0.262 N
Explanation:
It is given that,
Magnetic field of an electromagnet, B = 0.52 T
Length of the wire, 
Current in the straight wire, i = 10.5 A
Let F is the magnitude of force is exerted on the wire. The magnetic force acting on an object of length l is given by :
F = 0.262 N
So, the magnitude of force is exerted on the wire is 0.262 N.
