Ec₁=mv₁²/2=450*26²/2=225*676=152100 J
Ec₂=mv₂²/2=450*30²/2=225*900=202500 J
ΔEc=Ec₂-Ec₁=202500-152100=50.4 kJ
Answer:b)1770 kWh
Explanation:
Given
volume of water
Temperature rise
also 1 kg mass is approximately is 1 gallon
therefore 40,000 gallon is equivalent to 3.8\times 40000 kg
heat Required to raise temperature is
momentum= mass × velocity
p= 50×18
momentum= 900 kg m/s
a) To find the solution of this point we need to calculate the relation of the control resolution, that is,
Where
R= Range of the joint
B= Storage capacity
Making the substitution of the previous values we have,
B) For the now we need to calculate the accuracy, that is,
Where
= Standard deviation
The accuracy of the robot
Making the substitution,
c) At end we can to calculate the repeatability, that is,