G A 1.20-kg ball is tied to a 1.20-m long string is being spun in a vertical circle at a constant speed and with a period of 1.60 is. What is the maximum tension in the string

T = 17.42 N

Explanation:

angular velocity(ω) can be calculated from the expression:

=

Since T = 1.60

Then ω =

ω = 3.930 rad/s

Maximum tension of the string would be at the lowest point of the circle which can be written as:

T - mg = mrω ²

T = mg + mrω ²

T = m ( g + rω ²)

T = 1.2(9.8 + 1.20×3.930)

T = 17.4192 N

T = 17.42 N

  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

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