Across-country skier is skiing along at a zippy 8.0 m/s. she stops pushing and simply glides along, slowing to a reduced speed of 6.0 m/s after gliding for 5.0 m. what is the magnitude of her acceleration as she slows

The magnitude of her acceleration is -2.8 m/s²

Explanation:

It is given that,

Initial velocity of the skier, u = 8 m/s

Final velocity of the skier, v = 6 m/s

Distance covered, s = 5 m

We need to find the magnitude of acceleration as she slows down. It can be calculated using third equation of motion as :

The magnitude of her acceleration is -2.8 m/s²

Explanation:

It is given that,

Initial velocity of the skier, u = 8 m/s

Final velocity of the skier, v = 6 m/s

Distance covered, s = 5 m

We need to find the magnitude of acceleration as she slows down. It can be calculated using third equation of motion as :

So, the acceleration of the skier is -2.8 m/s². Hence, this is the required solution.

  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