To prevent damage to floors (and to increase friction) a crutch will often have a rubber tip attached to its end. If the end of the crutch is a circle of radius 0.95 cm without the tip, and the tip is a circle of radius 2.0cm, by what factor does the tip reduce the pressure exerted by the crutch

By a factor of about 0.23

Explanation:

Pressure is force over an area: P=F/A

Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that we can say:

P₁=(4/0.9025)P₂=4.4P₂   or

P₂=(0.9025/4)P₁=0.23P₁

What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.