Aparticle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0; (5.0 m, 5.0 m) at t = 2.0 s; and (14.0 m, 12.0 m) at t = 5.0 s. (a) find the magnitude of the average velocity from t = 0 to t = 2 s. (b) find the magnitude of the average velocity from t = 0 to t = 5 s.

a) Magnitude of the average velocity from t = 0 to t = 2 s = 2.24 m/s

b) Magnitude of the average velocity from t = 0 to t = 5 s = 3.16 m/s

Explanation:

a) Velocity is rate of change of position.

A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0

A particle's position coordinates (x, y) are (5.0 m, 5.0 m) at t = 2.0 s

Displacement = (5-1)i + (5-3)j = 4i + 2j

Change in time = 2s

Velocity

Magnitude of velocity

b) Velocity is rate of change of position.

A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0

A particle's position coordinates (x, y) are (14.0 m, 12.0 m) at t = 5.0 s

Displacement = (14-1)i + (12-3)j = 13i + 9j

Change in time = 5s

Velocity

Magnitude of velocity

  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

pe = weightxheight = 2000x1000=2,000,000joules