You decide to visit santa claus at the north pole to put in a good word about your splendid behavior throughout the year. while there, you notice that the elf sneezy, when hanging from a rope, produces a tension of 505 n in the ropea. if sneezy hangs from a similar rope while delivering presents at the earth's equator, what will the tension in it be? (recall that the earth is rotating about an axis through its north and south poles.)

To solve this problem it is necessary to apply the concepts related to the Rotational Force described from the equilibrium and Newton's second law.

When there is equilibrium, the Force generated by the tension is equivalent to the Force of the Weight. However in rotation, the Weight must be equivalent to the Centrifugal Force and the tension, in other words:

Where

Angular velocity is equal to the Period, at this case Earth's period

Radius of the Earth

m = mass

= Force of Tension

Newton's second law

Replacing and re-arrange to find the Tension we have,

Therefore when Sneezy is on the equator he is in a circular orbit with a Force of tension of 503.26N

  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

the answer for apex is the lowest