An open train car, with a mass of 2010 kg, coasts along a horizontal track at the speed 2.53 m/s. the car passes under a loading chute and, as it does so, gravel falls vertically into it for 3.25 s at the rate of 423 kg/s. what is the car's speed after the loading is completed? ignore rolling friction.

v' = 1.5 m/s

Explanation:

Given that,

Mass of the car, m = 2010 kg

Speed of the car, v = 2.53 m/s

The car passes under a loading chute and gravel falls vertically into it for 3.25 s at the rate of 423 kg/s. Mass loaded during this time, m' = 1374.75 kg

Let v' is the speed of car after the loading is completed. Using the conservation of momentum to find it.

v' = 1.5 m/s

So, the speed of the car after the loading is completed is 1.5 m/s. Hence, this is the required solution.

  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

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