Answer:
The correct option is;
Materials and Components
Explanation:
The efficiency of fluid power is influenced by the components and the materials used to deliver the power of the fluid as such fluid power control are focused on
1) Advances in fluid power
2) Making use of the advantages
3) Making use of the other externally available technological advantages
4) Giving allowance for disadvantages
Areas of interest in advances in fluid power are;
a. Computer optimized flow
b. The use of new and improved materials/coatings
c. The use of components that save energy, such as intelligent supply pressure adapting systems
Answer:
Explanation:
The detailed steps is as shown in the attachment.
Answer:
8861.75 m approximately 8862 m
Explanation:
We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.
for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that 
For simplicity we work with 
despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.
In accord with the formula 
the "normal" or "standard" weight of an object is given by 
when 
, so we need to find the value of 
that makes 
meaning that the original weight decrease by a 0.30%, so now we operate...
now we group like terms on the same sides 
we cancel equal tems on both sides and obtain that 
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
