Answer:
∆S1 = 0.5166kJ/K
∆S2 = 0.51826kJ/K
Explanation:
Check attachment for solution
Sorry man i don’t know sorry
Answer:
c = 18.0569 mm
Explanation:
Strategy
We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.
Given Data
Applied Torque
T = 750 N.m
Length of shaft
L = 1.2 m
Modulus of Rigidity
G = 77.2 GPa
Allowable Stress
г = 90 MPa
Maximum Angle of twist
∅=4°
∅=4*
/180
∅=69.813 *10^-3 rad
Required Diameter based on angle of twist
∅=TL/GJ
∅=TL/G*
/2*c^4
∅=2TL/G*
*c^4
c=
∅
c=18.0869 *10^-3 rad
Required Diameter based on shearing stress
г = T/J*c
г = [T/(J*
/2*c^4)]*c
г =[2T/(J*
*c^4)]*c
c=17.441*10^-3 rad
Minimum Radius Required
We will use larger of the two values
c= 18.0569 x 10^-3 m
c = 18.0569 mm
Answer:The answer is Potassium!
Explanation: This is true because each label should tell you about the available amount of a certain element. The standard order is Nitrogen-Phosphorus-Potassium. They are referred to by their standard abbreviations in the periodic table. One problem with fertilizer labels are that they are only required to disclose the amounts of macronutrients (or Nitrogen-Phosphorus-Potassium.)