If a 0.15 kg ball on the end of a string is swung in a vertical circle of radius .6 meters and makes 2 revolutions per second, what is the tension in the string at the very top of the circle? what is the tension in the string at the bottom of the circle? (12.7 n; 15.7 n)

12.7 N

15.7 N

Explanation:

mass (m) = 0.15 kg

radius (r) = 0.6 m

speed  = 2 rps = 2 x 60 = 120 rpm

acceleration due to gravity (g) = 9.8 m/s^{2}

find the tension at the top and bottom of the circle.

Tension at the top T =

        v =  radius x rpm x 0.10472 = 0.6 x 120 x 0.10472 = 7.54 m/s

         T = = 12.7 N      

Tension at the bottom T' =

        v =  radius x rpm x 0.10472 = 0.6 x 120 x 0.10472 = 7.54 m/s

         T' = = 15.7 N      

the second one and the forth one are in the purple because they are opinions and the rest are in the yellow

explanation:

  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