Question

Before digital storage devices, such as the memory in your cell phone, music was stored on vinyl disks with grooves with varying depths cut into the disk. A phonograph used a needle, which moved over the grooves, measuring the depth of the grooves. The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle with a 0.200-mm radius, what pressure is exerted on the record in Pa?

(a) ( 7.96 × 10⁵ , {Pa} )
(b) ( 3.18 × 10⁴ , {Pa} )
(c) ( 1.59 × 10⁶ , {Pa} )
(d) ( 6.36 × 10⁴ , {Pa} )

Answer 1

The pressure exerted by the phonograph needle on a vinyl record is calculated as 78.1 kPa, using the mass of the needle and the radius of its tip to determine the force and area, and then applying the pressure formula (Pressure = Force/Area).

Step-by-step explanation:

The calculation of the pressure exerted by a phonograph needle on a vinyl record involves understanding the relationship between force, area, and pressure. Pressure is defined as force per unit area. The force due to the weight of the needle is the mass times the acceleration due to gravity (F = mg). The area is the surface area of the needle's tip that contacts the record, which is a circle with a given radius.

First, convert the mass from grams to kilograms (1.00 g = 0.001 kg) because the SI unit for mass is kilogram. The force is then F = (0.001 kg)(9.8 m/s2) = 0.0098 N (newtons). Next, calculate the area of the circle using the formula A = πr2, where r is the radius in meters (0.200 mm = 0.0002 m). Thus, A = π(0.0002 m)2 = 1.256×10-7 m2. Finally, the pressure P is the force divided by the area, P = F/A.

Substitute the values into the pressure equation: P = 0.0098 N / 1.256×10-7 m2 = 7.81×104 N/m2 or 78.1 kPa, which is the pressure exerted on the record in pascals.