You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track to a height of 3.00 m.

how fast are you moving when you arrive at the 3.00-m height? - 22.1 m/s
- 20.8 m/s
- 23.4 m/s
- 14.7 m/s

To solve this problem, we need to use both energy conservation and potential kinetic equations.

When the energy accumulated from a certain height is released, it becomes 'motion' energy or kinetic energy. Mathematically this can be expressed as

PE = KE

Where

m = mass

g = Gravitational acceleration

v = Velocity

Change of the height

We know that the body, based on a reference system where the floor is the zero coordinate, starts from being 25 meters high to fall to 3 meters high, so the total difference in height would be

We also have to

Using the previous equation we have to:

Replacing

The correct answer is 20.8m/s

explanation:

a fuel engine ( known as a gasoline engine in american english )is an internal combustion engine with spark ignition, designed to run on petrol (gasoline) and similar volatile fuels.

  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