A165g arrow is shot straight up into the air with a speed of 16m/s reaches a maximum height of 12.0612m.

a. find the initial kinetic energy. the acceleration of gravity is 9.8m/s^2.
b. find the potential energy at its highest position.
c. find the magnitude of the energy lost due to air resistance.

a. 21.12 J

b. 19.50 J

c. 1.62 J

Explanation:

Kinetic Energy: this is the energy of a body due to motion.

it is represented mathematically as

Ek = 1/2mv² Equation 1

Where Ek = kinetic energy, m = mass of the arrow, v = velocity of the arrow

Potential Energy: This is the energy of a body due to position.

It is represented mathematically as

EP = mgh Equation 2

Ep = potential Energy, h = height, g = acceleration due to gravity.

(a) The initial kinetic energy.

using Equation 1,

Ek = 1/2mv²

given: m = 165 g = 0.165 kg, v = 16 m/s.

Substituting these values into equation 1

Ek = 1/2(0.165)(16)²

Ek = 21.12 J.

(b) The potential energy at its highest position

Using Equation 2

Ep = mgh

given: m = 165 g = 0.165 kg, h = 12.0612 m, g = 9.8 m/s²

Substituting these values into equation 2

Ep = 0.165(12.0612)(9.8)

Ep = 19.50 J

(c) Energy lost due to air resistance

= Initial kinetic energy - potential energy at the highest position

EL = Ek - Ep

Where EL = energy lost.

EL = 21.12 - 19.50

EL = 1.62 J

to treat tumors in the body

  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