A roller coaster has a mass of 1200.0kg. The coaster is going 22.0 m/s at the bottom of the third loop-the-loop that is 2.5m above the ground. Determine the height of
the first hill that is required, assuming the cart is stationary at the top of the first hill
before it falls.

h = 27.17 m

Explanation:

First, we will calculate the total mechanical energy of the system at the bottom point of the third loop:

Mechanical Energy = Kinetic Energy + Potential Energy

where,

E = Total Mechanical Energy = ?

m = mass of the roller coaster = 1200 kg

v = velocity of the roller coaster = 22 m/s

g = acceleration due to gravity = 9.81 m/s²

h = height of roller coaster = 2.5 m

Therefore,

Now, the total mechanical energy at the top position of the first hill must also be the same:

where,

v = 0 m/s

h = ?

Therefore,

h = 27.17 m

wavelength variability is the right one

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz