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g A computer is reading data from a rotating CD-ROM. At a point that is 0.0189 m from the center of the disk, the centripetal acceleration is 241 m/s2. What is the centripetal acceleration at a point that is 0.0897 m from the center of the disc?

Sagot :

nuhulawal20

Answer:

the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/sĀ²

Explanation:

Given the data in the question;

centripetal acceleration a[tex]_c[/tex]ā‚ = 241 m/sĀ²

radius rā‚ = 0.0189 m

radius rā‚‚ = 0.0897 m

centripetal acceleration a[tex]_c[/tex]ā‚‚ = ? m/sĀ²

since the rotational period will be the same for the two disk,

we use the centripetal acceleration formula a[tex]_c[/tex] = (4Ļ€Ā²r/TĀ²) to find the rotational period for the first disk.

a[tex]_c[/tex]ā‚ = (4Ļ€Ā²rā‚/TĀ²)

make TĀ² subject of formula

TĀ² = 4Ļ€Ā²rā‚ / a[tex]_c[/tex]ā‚

we substitute

TĀ² = ( 4 Ɨ Ļ€Ā² Ɨ 0.0189 ) Ā / 241 Ā 

TĀ² = 0.00309602528 sĀ²

Now we use the same formula to find a[tex]_c[/tex]ā‚‚

a[tex]_c[/tex]ā‚‚ = ( 4Ļ€Ā²rā‚‚ / TĀ² )

we substitute

a[tex]_c[/tex]ā‚‚ = ( 4 Ɨ Ļ€Ā² Ɨ 0.0897 ) Ā / 0.00309602528

a[tex]_c[/tex]ā‚‚ = 1143.8 m/sĀ²

Therefore, the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/sĀ²

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