Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 78 m long. the coefficient of kinetic friction between rubber and the pavement is about 0.90. estimate the initial speed of that car assuming a level road.

37.11231 m/s

Explanation:

u = Initial velocity

v = Final velocity = 0

s = Displacement = 78 m

g = Acceleration due to gravity = 9.81 m/s²

= Coefficient of kinetic friction = 0.9

Acceleration is given by

Equation of motion

The initial speed of that car is 37.11231 m/s

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

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