Question

the force on phonogram needle is 1.2 N. The point has a circular cross-section of radius 0.1 mm. What pressure does it exert on the record in (1) Pa (2) atm​

Answer 1

answer : 1)387*10^5 pa 2) 381.939304atm

Explanation:

pressure ( pa) = force ( newton)/area (m^2)

0.1mm = 0.1*10^-6 pa

1 .2 Newtons /0.1*10^-6

=387*10^5 pa

1atm = 1.013 ×10^5 pa

387*10^5 pa= 381.939304atm

answer : 1)387*10^5 pa 2) 381.939304atm

hope it will help you

Answer 2

Answer:

Explanation:

Here, forceo n the phonogram needle, F=1.2N

radius of the circular needle, r=0.1mm=0.1×10−3m

area of cross-section of the needle, A=πr2

=3.142×(0.1×10−3)2=3.142×10−8m2

pressure exerted on the record in Pa. P=PA

=(1.2N)(3.142×10−8m2)=382×105Nm□

pressure exerted on the record in atm

382×105Nm21.013×105Nm2atm

=378atm

hope it helpful!