Three arrows are shot horizontally. They have left the bow and are traveling parallel to the ground. Air resistance is negligible. Rank in order, from largest to smallest, the magnitudes of the horizontal forces F1, F2, and F3 acting on the arrows. Some may be equal. State your reasoning.

Arrow 1: is 80g @ 10 m/s
Arrow 2: is 80g @ 9 m/s
Arrow 3: is 90g @ 9 m/s

The third arrow has the greatest force, followed by the first arrow and the least is the second arrow. (F₃ > F₁ > F₂)

The given parameters;

The magnitude of the horizontal force on each arrow is determined by applying Newton's second law of motion;

The magnitude of force on an object is directly proportional to the velocity but inversely proportional to time action.

For example:

The magnitude of the horizontal force on each arrow is calculated as;

Thus, we can conclude that the third arrow has the greatest force, followed by the first arrow and the least is the second arrow. (F₃ > F₁ > F₂)

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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