Achild is sitting on the seat of a swing with ropes 10 m long. her father pulls the swing back until the ropes make a 47º angle with the vertical and then releases the swing. if air resistance is neglected, what is the speed of the child at the bottom of the arc of the swing when the ropes are vertical?

7.9 m/s

Explanation:

A right triangle is formed where one leg (let's call it x) form an angle of 47° with the hypotenuse and the hypotenuse is the rope. From cosine definition:

cos 47° = x/10

x = 10*cos 47°

x = 6.82 m

The difference between the rope length and this value gives us the height of the child before her father releases the swing. Let's call this height h.

h = 10 m - 6.82 m = 3.18 m

Before the swing starts the child is not moving, then all of her energy is potential energy (PE). When she gets at the bottom of the arc of the swing,  all of her PE is transformed into kinetic energy (KE).

KE = PE

(1/2)*m*v² = m*g*h

where v is her speed and g is the standard gravitational acceleration. Solving for v:

v = √(2*g*h)

v = √(2*9.81*3.18)

v = 7.9 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