Answer:
<u>Assistants</u><u> </u><u>works alongside and assists the engineers.</u>
“Thinking about pleasant things to pass the time” would not promote safety in the shop because it would be taking the focus away from important tasks, which in turn decreases safety.
Answer:
98°C
Explanation:
Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7
22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²
Temperature change, t = (50 - 25)°C = 25°C = 298K
Hence, Temperature = 150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =
∴ Temperature change = 2.00K
But temperature, T= (373 - 2)K = 371 K
In °C = (371 - 273)K = 98°C
Answer:
radius = 9.1 ×
m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius =
.................1
here F is applied load and is length
put here value and we get
radius =
solve it we get
radius = 9.1 ×
m
Answer:
b) The null hypothesis should be rejected.
Explanation:
The null hypothesis is that the mean shear strength of spot welds is at least
3.1 MPa
H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa
The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.
This is one tailed test
The critical region Z(0.05) < ± 1.645
The Sample mean= x`= 3.07
The number of welds= n= 15
Standard Deviation= s= 0.069
Applying z test
z= x`-u/s/√n
z= 3.07-3.1/0.069/√15
z= -0.03/0.0178
z= -1.68
As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa