(i) a 0.145-kg baseball pitched at 31.0 m/s is hit on a horizontal line drive straight back at the pitcher at 46.0 m/s. if the contact time between bat and ball is 5.00 × 10^–3 s, calculate the force (assumed to be constant) between the ball and bat.

The force between the ball and the bat = 2233 N

Explanation:

Force: This can be defined as the product of force and its acceleration.

The S.I unit of Force is Newton (N)

F = ma Equation 1

Where F = Force , m = mass of the ball, a = acceleration of the ball.

where

a = (v-u)/t Equation 2

Where v = final velocity of the ball, u = initial velocity of the ball, t = time of contact between the ball an the bat.

Given: v = 46 m/s u = - 31 m/s (it hit an horizontal line drive back at the pitcher), t = 5×10⁻³ s

Substituting these values into equation 2,

a = [46-(31)]/(5×10⁻³ )

a = 77×10³/5

a = 15.4×10³ m/s².

Also given m = 0.145 kg

Substituting into equation 1,

F = 0.145(15.4×10³ )

F = 2233 N

Thus the force between the ball and the bat = 2233 N

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  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