Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
Answer:
<em><u>The 'shoulder' of a road is the land to the edge of the road. On most roads without pavements, the shoulder is a strip of grass or a hedgerow. This is known as a 'soft shoulder'. On a motorway, this strip of land is hardstanding, hence the name 'hard shoulder.'</u></em>
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Answer:
Aqueous solution of ionic compounds conduct electricity while solid ionic compounds don't.
Explanation:
Ionic compound conduct electricity when liquid or in aqueous solution that is resolved in water because the ionic bonds of the compound become weak and the ions are free to move from place to place.
Ionic compounds don't conduct electricity while in solid state because the ionic bonds are to strong and ions cannot move around with lack of space for movement which makes the electric conductivity zero.
Explanation:
The end-use industries of thermochromic materials include packaging, printing & coating, medical, textile, industrial, and others. Printing & coating is the fastest-growing end-use industry of thermochromic materials owing to a significant increase in the demand for thermal paper for POS systems. The use of thermochromic materials is gaining momentum for interactive packaging that encourages consumers to take a product off the shelf and use it.
Answer:

Explanation:
We are given:
m = 1.06Kg

T = 22kj
Therefore we need to find coefficient performance or the cycle


= 5
For the amount of heat absorbed:

= 5 × 22 = 110KJ
For the amount of heat rejected:

= 110 + 22 = 132KJ
[tex[ q_H = \frac{Q_L}{m} [/tex];
= 
= 124.5KJ
Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c
Convert 69.5°c to K we have 342.5K
To find the minimum temperature:
;

= 285.4K
Convert to °C we have 12.4°C
From the refrigerant R -134a table at
= 12.4°c we have 442KPa