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  1. Solution: V = 26.3 m/s Method: final KE = work done by (pitcher + gravity) + initial KE work done by pitcher Wp = F*d = F*πr = 30.8N*π*0.594m = 57.5 J Wg = mgh = mg*2r = 0.244kg * 9.8m/s² * 2*0.594m = 2.84 J...

    1 Answers · Science & Mathematics · 03/01/2021

  2. ... level, what is its potential energy when the textbook is released ? PE = mgh = 1.60(9.81)(10.0) = 156.96 ≈ 157 J If the gravitational...

    1 Answers · Science & Mathematics · 05/01/2021

  3. ... for a moment, B static at the dip down (ie a not this question) If released , B will roll down the dip, along the flat bit and up to the matching height...

    3 Answers · Science & Mathematics · 13/01/2021

  4. No. The court could not prove whether the cat did, or did not, eat the poison. Schrodinger was released on the basis of uncertainty.

    5 Answers · Science & Mathematics · 05/01/2021

  5. 1. This question is actually a bit trickier than it seems at first because of the collision between the bullet and the block. If the bullet is embedded into the block, it's clear that there is deformation so we have to consider energy losses in the...

    3 Answers · Science & Mathematics · 31/12/2020

  6. depth (d) of the well is given by two equations. Ignoring air resistance, the stone drops d = ½(9.8)ts² The sound travels the same distance in time 5 - ts d = 340(5 - ts) ½(9.8)ts² = 340(5...

    1 Answers · Science & Mathematics · 12/01/2021

  7. range formula when launch and impact height are the same: x = V²sin(2Θ)/g so here 73.0 m = (37.0m/s)² * sin(2Θ) / 9.81m/s² sin(2Θ) = 0.523 2Θ = 31.5º Θ = 15.8º Hope this helps!

    1 Answers · Science & Mathematics · 14/01/2021

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