Answer:
Due to touch of teflon, its charge will reduce but will not go to zero. Some amount of its initial charge will be transferred to Aluminum rod. So, aluminum rod will have a non-zero negative charge.
Explanation:
Answer:
At a retirement party, a coworker described Terry as dedicated, hardworking, and dependable. He also said that Terry was a great leader, knew the computer system, and kept the company's finances in order
Answer:
The temperature of the strip as it exits the furnace is 819.15 °C
Explanation:
The characteristic length of the strip is given by;
The Biot number is given as;
< 0.1, thus apply lumped system approximation to determine the constant time for the process;
The time for the heating process is given as;
Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;
Therefore, the temperature of the strip as it exits the furnace is 819.15 °C
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
