Astudent, starting from rest, slides down a water slide. on the way down, a kinetic frictional force (a nonconservative force) acts on her. the student has a mass of 77 kg, and the height of the water slide is 12.0 m. if the kinetic frictional force does -6.7 × 103 j of work, how fast is the student going at the bottom of the slide?

7.8 m/s

Explanation:

At the top of the slide, the mechanical energy of the student is just gravitational potential energy, given by:

where m=77 kg is the mass of the student, g=9.8 m/s^2 is the gravitational acceleration, h=12.0 m is the height of the slide. Substituting,

While she slides down, the frictional force does

on her. So, the final mechanical energy of the student at the bottom of the slide is

And this energy is just equal to the final kinetic energy of the student, since its potential energy is just zero (at the bottom of the slide, h=0):

where v is the final speed of the student. Solving for v,

  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