A20.0-m-tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. a strong wind bends the pole as much as a horizontal 900.0-n force on the top would do. how far to the side does the top of the pole flex?

To solve the problem it is necessary to apply the concepts related to deflection in a metal.

By definition the deflection is given by the equation

Where

Force

A =Area

s = shear modulus

= Original length

To calculate the area we must obtain the radius generated by the deflection, that is to say

r = \frac{D}{2}

Where,

D = Diameter

r = Radius of the aluminum flagpole

Therefore the cross sectional area of the pole is given as

Replacing our values at the previous equation we have that

Therefore the deflection in the pole is 0.57mm

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia