## anonymous 4 years ago 4) Eight million (8,000,000) kg of water are at the verge of dropping over Niagara Falls to the rocks 50.0 meters below. What is the change in gravitational potential energy as the water splashes on the rocks below?

1. anonymous

U=mgh $\Delta U=mgh_1-mgh_0=mg(h_1-h_0)=mg( \Delta h)$ $\Delta U = 8,000,000*9.8*(-50)$

2. anonymous

thanks

3. anonymous

$\Delta=\text{'change \in'}$

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