Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.
3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.
Answer:
Caustic soda would be least affected by a gas curtailment
Explanation:
Let,
X1= ammonia
X2= ammonium phosphate
X3= ammonium nitrate
X4 = Urea
X5= hydro-fluoric acid
X6= Chlorine
X7= Caustic soda
X8= Vinyl chloride monomer.
Agri chem’s current natural gas usage
= (1200 × 8 + 540 × 10 + 490 × 12 + …)
= 85,680,000 cu. ft. per day
When, the curtailment is 20%, availability is
= 0.8 X 85,680
= 68,554,000 cu. ft. per day
Therefore, the gas constant
= 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8 ≤68,544
When, the curtailment is 40% availability is
= 0.6 X 85,680
= 51,408,000 cu. ft. per day
Constant = 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8≤ 51,408
