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cutie.patootie

  • 3 years ago

A 5.5 kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75 s later. Find the net external force on the falling ball.

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  1. yrelhan4
    • 3 years ago
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    s=1/2at^2 you have s and t find a. net force=ma. plug and solve.

  2. cutie.patootie
    • 3 years ago
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    What is s?

  3. yrelhan4
    • 3 years ago
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    s-->distance travelled a--> acceleration due to gravity here t-->time taken

  4. cutie.patootie
    • 3 years ago
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    Okay, thanks:)

  5. yrelhan4
    • 3 years ago
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    cheers!

  6. cutie.patootie
    • 3 years ago
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    I 431.2N but that isn't an option..

  7. yrelhan4
    • 3 years ago
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    43.12 N is the answer.

  8. cutie.patootie
    • 3 years ago
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    So this is what I did.. s=12m a=? t=1.75s 12=((1.75)^2a)/2 12=(3.0625a)/2 24=3.0625a a=7.84 netforce=55*7.84 netforce=431.2N

  9. yrelhan4
    • 3 years ago
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    Net acceleration. a = 2*12/1.75² Net force ma = 5.5* 2*12/1.75² = 43.12 N

  10. cutie.patootie
    • 3 years ago
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    I get what your saying now. But the only problem is my choices are 120N, 54N, & 0N. But I can rule out 0N automatically

  11. yrelhan4
    • 3 years ago
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    hmm. no idea. sorry.

  12. cutie.patootie
    • 3 years ago
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    Its okay, thanks for your help though :) I really do appreciate it

  13. yrelhan4
    • 3 years ago
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    you're welcome. good luck. :)

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