Asmall metal ball is suspended from the ceiling by a thread of negligible mass. the ball is then set in motion in a horizontal circle so that the thread describes a cone. the acceleration of gravity is 9.8 m/s2, 2.7 m, 6 kg, 31c. what is the speed of the ball when it is in circular motion? answer in units of m/s.

a) -2.1731 m⁻s

Explanation:

L= length of thread, H= height of ball from ceiling,

N/B h= l cosФ, Ф=31°, l=2.7m,  g=9.8m⁻s, m= 6kg

Required to find speed v, of the ball in m⁻s

ω=√g/h, =√g/lcosФ

⇒ ω=√9.8/2.7×cos31°

= 1.992rad⁻s

Now, speed in m/s, v= r.ω

where tanФ= r/h ⇒

r= h×cos 31°

r= 2.7 × cos(31) × tan(31) =-1.0909m

v= 1.992× -1.0909 =  -2.1731 m/s

  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