Answer:
Total Work=2275000 ft-lb
Explanation:
According to Riemann sum approximate for work needed to lift the cable:
Sine we have to add 650 terms because distance 650 ft, we will us the integration.
![W=\int\limits^a_b {f(x)} \, dx \\W=\int\limits^{650}_0 {8x} \, dx \\W=[4x^2]_0^{650}\\W=4(650)^2-0^2\\W=4*422500 ft-lb\\W=1690000 ft-lb](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5Ea_b%20%7Bf%28x%29%7D%20%5C%2C%20dx%20%5C%5CW%3D%5Cint%5Climits%5E%7B650%7D_0%20%7B8x%7D%20%5C%2C%20dx%20%5C%5CW%3D%5B4x%5E2%5D_0%5E%7B650%7D%5C%5CW%3D4%28650%29%5E2-0%5E2%5C%5CW%3D4%2A422500%20ft-lb%5C%5CW%3D1690000%20ft-lb)
Work done on lifting:
Total Work= 
Total Work=1690000+585000
Total Work=2275000 ft-lb
Answer:
Explanation:
Let assume that pole vaulter begins running at a height of zero. The pole vaulter is modelled after the Principle of Energy Conservation:
The expression is simplified and final height is cleared within the equation:
![h_{B} = \frac{[(11\,\frac{m}{s} )^{2}-(1.3\,\frac{m}{s} )^{2}]}{2\cdot (9.807\,\frac{m}{s^{2}} )}](https://tex.z-dn.net/?f=h_%7BB%7D%20%3D%20%5Cfrac%7B%5B%2811%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D-%281.3%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%5D%7D%7B2%5Ccdot%20%289.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%29%7D)
Answer:
the answer for this is my.
It was in Texas on September 8, 1900.
