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valentinak56 [21]
3 years ago
6

A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block is suspended from the free end of the string.

If the block is released from rest at a distance of 2.80 m above the floor, what is its speed just before it strikes the floor
Physics
1 answer:
LenaWriter [7]3 years ago
7 0

Answer: 7.41 m/s

Explanation: By using the law of of energy, kinetic energy of the brick as it falls equals the potential energy before falling.

Kinetic energy = mv²/2, potential energy = mgh

mv²/2 = mgh

v²/2 = gh

v² = 2gh

v = √2gh

Where g = 9.8 m/s², h = 2.80m

v = √2×9.8×2.8 = 7.41 m/s

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Since the guitar player wants to hear 9 beats in 2 seconds, then:

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Then, the pitches that can be used are those whose difference with respect to the frequency of 343Hz is 4.5Hz:

\begin{gathered} f_2=f_1\pm f_{beat} \\  \\ \Rightarrow f_2=343Hz\pm4.5Hz \\  \\ \therefore f_{2,1}=347.5Hz \\ f_{2,2}=338.5Hz \end{gathered}

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