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Sagot :
Explanation:
Length of a string, l = 0.626 m
A tennis ball revolves in a horizontal circle above his head at a rate of 4.50 rev/s.
(a) The centripetal acceleration of the tennis ball is given by :
[tex]a=\omega^2r\\\\a=(4.5\times 2\pi)^2\times 0.626\\\\=500.44\ m/s^2[/tex]
(b) When r = 0.626 m and ω = 4.50 rev/s
Speed,
[tex]v=r\omega\\\\=0.626\times 4.50 \times 2\pi\\\\=17.69\ m/s[/tex]
When r = 1 m and ω = 4 rev/s
Speed,
[tex]v=r\omega\\\\=1\times 4 \times 2\pi\\\\=25.13\ m/s[/tex]
Speed is more in second case when child now increases the length of the string to 1.00m but has to decrease the rate of rotation to 4.00 rev/s.
(c) [tex]a=\omega^2r\\\\a=(4\times 2\pi)^2\times 1\\\\=631.65\ m/s^2[/tex]
Hence, this is the required solution.
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