For this quiz, we shall return to the radio control car track that we visited briefly on the last quiz. The track is 10 meters long and perfectly straight. A series of reference marks are 1.0 meter apart along the track. A judge sets her stopwatch to 0.0 seconds, then she starts her watch at the instant the car passes the 2.0 meter mark. When the car passes the 8.0 meter mark, the judge reads 3.9 seconds on her stopwatch. Using equation x:=:x0:+:vt x = x 0 + v t , calculate v v in meters per second.

Explanation:

The given equation is where denotes initial time in the stopwatch and denotes the distance covered by car.

As the judge starts her watch at the instant the car passes the 2.0 meter mark   and the judge reads 3.9 seconds on her stopwatch when the car passes the 8.0 meter mark.

Put in equation

  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

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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explanation: