A 0.5 kg wrench is dropped from a height of 10 meters. Using v = square root of2gh, where g = 9.80 m/s^2, what is the momentum just before it strikes the floor?
"mgh= \\frac{1}{2}mv^2 \\\\\nv= \\sqrt{2gh} \\\\\n= \\sqrt{2 \\times 9.8 \\times 10} \\\\\n= 14.0 \\;m\/s \\\\\np =mv \\\\\n= 0.5 \\times 14 \\\\\n= 7.0 \\;kg \\;m\/s, \\;dowm"
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