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Kuroks

  • 3 years ago

The Steamboat aloftser in Yellowstone National Park, Wyoming is capable of shooting its hot water up from the ground with a speed of 48.0 m/s. How high can this geyser shoot? (HINT: USE KINEMATIC EQUATIONS)

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  1. henkjan
    • 3 years ago
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    use energy conservation kinetic energy is converted into potential here. Ekin = 1/2mv^2 Epot =m*g*h 1/2V^2 = g*h

  2. Kuroks
    • 3 years ago
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    Im confused, can you explain it more?

  3. henkjan
    • 3 years ago
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    You conserve energy here..... the kinetic energy in the start is 1/2*m*v^2 ALL this energy is converted to potential energy. The potential energy at the end is given as: m*g*h You know they are equal, so we can use an equation to calculate the height h. m*g*h = m*1/2*V^2 We see that we can take m out: g*h = 1/2 V^2

  4. Kuroks
    • 3 years ago
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    So you cant do anything other than that?

  5. henkjan
    • 3 years ago
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    Yes, you can... but this is the easiest way to do it. You can also solve by using s = 1/2*a*t^2.. a = g = 9.81 m/s2 t = v/g = 48/9.81 s = 1/2 * a * (v/g)^2 = 9.81*(48/9.81)^2

  6. henkjan
    • 3 years ago
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    But it boils down into the same equation as: s = h = 1/2 * g * (v/g)^2 = 1/2 *V^2/g

  7. Kuroks
    • 3 years ago
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    Okay, thank you ^^

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