Question

The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 0.600 g is supported by a needle, the tip of which is a circle 0.240 mm in radius, what pressure is exerted on the record in N/m2?

Answer 1

`P=3.25x10^(4)(N)/(m^2)`

Step-by-step explanation:

Hello,

In this case, since pressure is defined as the force applied over a surface:

`P=(F)/(A)`

We can associate the force with the weight of the needle computed by using the acceleration of the gravity:

`F=0.600g*(1kg)/(1000g)*9.8(m)/(s^2) =5.88x10^(-3)N`

And the area of the the tip (circle) in meters:

`A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*(1m)/(1000mm) )^2\\\\A=1.81x10^(-7)m^2`

Thus, the pressure exerted on the record turns out:

`P=(5.88x10^(-3)N)/(1.81x10^(-7)m^2) \\\\P=3.25x10^(4)(N)/(m^2)`

Which is truly a large value due to the tiny area on which the pressure is exerted.

Best regards.