Answer:
kinematic viscosity is 0.0149 ft²/s
Explanation:
given data
specific gravity S = 0.94
density ρ = 0.94 × 1000
viscosity μ = 13 Poise = 1.3 Pa-sec
we know 1 poise = 0.1 pas
to find out
kinematic viscosity
solution
we will apply here Kinematic viscosity formula that is
kinematic viscosity = 
..................1
put here value in equation 1
and here ρ is density and μ is viscosity
kinematic viscosity = 
kinematic viscosity = 1.382978 × 
m³/s
so kinematic viscosity is 0.0149 ft²/s
Answer: False
Explanation: For implementation of the safety and the health program, there is high requirement of the employees because they are the people who will participate in the program and can make it successful.
Apart from the management and the leader, who are there to manage the program and lead the program as head by taking responsibility of other respectively, employees participation is required.Employees are referred as the people who will do the activities and also teach other about it.Thus, the statement in question is false.
Answer:
5.192
Explanation:
This is a two tailed test hence the null and alternative hypothesis will be
Sample size, n=674
Sample proportion 
Z test statistic formula for one sample is given by
Substituting the values given then
Answer:
The elastic modulus of the steel is 139062.5 N/in^2
Explanation:
Elastic modulus = stress ÷ strain
Load = 89,000 N
Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2
Stress = load/area = 89,000/0.64 = 139.0625 N/in^2
Length of steel bar = 4 in
Extension = 4×10^-3 in
Strain = extension/length = 4×10^-3/4 = 1×10^-3
Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2
