The distance required for a car to come to a stop will vary depending on how fast the car is moving. Suppose that a certain car traveling down the road at a speed of 20 m / s 20 m/s can come to a complete stop within a distance of 80 m 80 m . Assuming the road conditions remain the same, what would be the stopping distance required for the same car if it were moving at speeds of 10 m / s 10 m/s , 40 m / s 40 m/s , or 5 m / s 5 m/s ?

Explanation:

If F be the force acting on the car to stop it

work done by force = change in kinetic energy

F X 80 = 1/2 m x 20²

F = 2.5 m

If speed is 10m/s

for distance d , the equation is

F x d = 1/2 x m x 10²

2.5 m x d = 1/2 x m x 10²

d = 20 m

If velocity is 40 m /s

F x d = 1/2 x m x 40²

2.5m x d = 1/2 m x 1600

d = 320 m

For v = 5 m/s

F x d = 1/2 m x 5²

2.5 m x d = 1/2 m x 5²

d = 5 m .

it becomes positively charged by charging it

  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