in a simple machine the input work is 187.5 N and output work is 125N. If load distance is 25 cm and effort distance is 75 cm calculate value of load and effort ​

1. Laod = 500 N

2. Effort = 250 N

Explanation:

From the question given above, the following data were obtained:

Input work = 187.5 J

Output work = 125 J

Load distance = 25 cm

Effort distance = 75 cm

1. Determination of the load.

Output work = 125 J

Load distance = 25 cm = 25/100 = 0.25 m

Load =?

Output work = Load × Load distance

125 = Load × 0.25

Divide both side by 0.25

Load = 125 / 0.25

Laod = 500 N

2. Determination of the effort.

Input work = 187.5 J

Effort distance = 75 cm = 75/100 = 0.75 m

Effort =?

Input work = Effort × Effort distance

187.5 = Effort × 0.75

Divide both side by 0.75

Effort = 187.5 / 0.75

Effort = 250 N

  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

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