Abowling ball encounters a 0.760 m vertical rise on the way back to the ball rack. ignore frictional losses and assume the mass of the ball is distributed uniformly. the translational speed pf the ball is 3.5 m/s at the bottom of the rise. find the translational speed at the top. note that moment of inertia of a solid sphere about an axis tanget to its surface is i = (7/5) mr^2

To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

Where,

V = linear speed

R = the radius

In the case of the initial kinetic energy:

Where I is the moment of inertia previously defined.

In the case of the final kinetic energy, we have to,

For conservation of Energy we have, that

, then (canceling the mass and the radius)

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

the part of the wave he is travelling on is the crest