During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leaping the river with his car. The side of the road the car is on is18.0 m above the river, while the opposite side is only1.8 m above the river. The river itself is a raging torrent69.0 m wide. Part AHow fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side?Part BWhat is the speed of the car just before it lands on the other side?

Part A: The speed the car should be travelling when leaping the river is approximately 37.948 m/s

Part B: The speed of the car just before it lands is approximately 41.92345 m/s

Explanation:

The parameters of the car attempting leaping the river are;

The height of the car over the river = 18 m

The height of the opposite side of the bridge above the river = 1.8 m

The width of the river, x = 69.0 m

Part A

The time, 't' it would take the car to fall from 18 m above the river to 1.8 m above the river is given as follows;

t = √(2·h/g)

Where;

h = The height of the fall = 18 m - 1.8 m = 16.2 m

g = The acceleration due to gravity = 9.8 m/s²

∴ t = √(2×16.2 m/(9.8 m/s²)) = (9/7)·√2 s

The horizontal speed, 'vₓ', with which the car should be travelling at the time it leaves the road in order just to clear the river and land safely on the opposite side is given as follows;

vₓ = x/t = 69.0 m/((9/7)·√2 s) = (161/6)·√2 m/s ≈ 37.948 m/s

The horizontal speed the car should be travelling when leaping the river, vₓ ≈ 37.948 m/s

Part B;

The vertical velocity of the car is given as follows;

² = ² + 2·g·h

∴ ² =2·g·h = 2 × 9.8 m/s² × 16.200 m = 317.52m²/s²

= √(317.52 m²/s²) = (63/5)·√2 ≈ 17.819 m/s

The magnitude of the speed of the car, 'v', just before it lands is given using Pythagoras' theorem for resultant vectors as follows;

v = √(² + vₓ²) = √(317.52 m²/s² +((161/6)·√2 m/s)²) ≈ 41.92345 m/s

The speed of the car just before it lands, v ≈ 41.92345 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

prolly a beacuse the momentuem the car has increases while going down hill