An open train car, with a mass of 2150 kg, coasts along a horizontal track at the speed 2.83 m/s. The car passes under a loading chute and, as it does so, gravel falls vertically into it for 2.95 s at the rate of 481 kg/s. What is the car's speed v f after the loading is completed? Ignore rolling friction.

The car's final speed after the loading is completed is 1.71 m/s

Explanation:

Given;

mass of the open car, m₁ = 2150 kg

initial speed of the car, u = 2.83 m/s

loading rate of the gravel, m' = 481 kg/s.

loading time of the gravel, t = 2.95 s

Then, mass of the gravel, m₂ = m' x  t  = 481 x 2.95 = 1418.95 kg

To determine the car's speed after the loading is completed, we apply the principle of conservation of linear momentum;

Initial momentum = Final momentum

m₁u = v(m₁ + m₂)

where;

v is the car's final speed after the loading is completed

2150 x 2.83 = v (2150 +  1418.95)

6084.5 = 3568.95v

v = 6084.5 / 3568.95

v = 1.71 m/s

Thus, the car's final speed after the loading is completed is 1.71 m/s

  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

this could be newton#'s law of cooling. the cooling curve is exponential