Answer:
i) The torque required to raise the load is 15.85 N*m
ii) The torque required to lower the load is 6.91 N*m
iii) The minimum coefficient of friction is -0.016
Explanation:
Given:
dm = mean diameter = 0.03 m
p = pitch = 0.004 m
n = number of starts = 1
The lead is:
L = n * p = 1 * 0.004 = 0.004 m
F = load = 7000 N
dc = collar diameter = 0.035 m
u = 0.05
i) The helix angle is:
The torque is:
ii) The torque to lowering the load is:
iii)
Clearing u:
u = -0.016
Answer:
i) SF: 
ii) BM : 
Explanation:
Let's take,
Making y the subject of formula, we have :
For shear force (SF), we have:
This is the area of the diagram.
The shear force equation =
For bending moment (BM):
The bending moment equation =
Answer:
(iv) second law of thermodynamics
Explanation:
The Clausius inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero
AnswerWhat Are the Classifications of Burns? Burns are classified as first-, second-, or third-degree, depending on how deep and severe they penetrate the skin's surface. First-degree burns affect only the epidermis, or outer layer of skin. The burn site is red, painful, dry, and with no blisters.
Explanation:
