If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?

Physics

1 answer:

mamaluj [8]3 years ago

4 0

Answer:

Explanation:

$KE=\frac{1}{2}mv^2$

$784=\frac{1}{2}(40)v^2$

$784=20v^2$

$39.2=v^2$

$v=6.26m/s$

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Answer:

a) $f_1=5.587Hz$

b) $f_{n+1}-f_n=5.587Hz$

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The frequency of the $n^{th}$ harmonic of a vibrating string of length L, linear density $\mu$ under a tension T is given by the formula:

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b) The frequency difference between successive modes is the fundamental frequency, since:

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