Answer:
Overall ideal mechanical advantage of the machine = 40
Explanation:
Given:
Ideal mechanical advantage of three machine = 2, 4, 5
Find:
Overall ideal mechanical advantage of the machine
Computation:
Overall ideal mechanical advantage of the machine = 2 × 4× 5
Overall ideal mechanical advantage of the machine = 40
Answer:
Correct option: B. 90%
Explanation:
The confidence interval is given by:
![CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]](https://tex.z-dn.net/?f=CI%20%3D%20%5B%5Cbar%7Bx%7D%20-%20z%5Csigma_%7B%5Cbar%7Bx%7D%7D%20%2C%20%5Cbar%7Bx%7D%2Bz%5Csigma_%7B%5Cbar%7Bx%7D%7D%20%5D)
If
is 190, we can find the value of
:



Now we need to find the value of
:


So the value of z is 1.71.
Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436
This value will occur in both sides of the normal curve, so the confidence level is:

The nearest CI in the options is 90%, so the correct option is B.
Answer:
1. 
2. 
Explanation:
1.
Given:
- height of the window pane,

- width of the window pane,

- thickness of the pane,

- thermal conductivity of the glass pane,

- temperature of the inner surface,

- temperature of the outer surface,

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

here:
A = area through which the heat transfer occurs = 
dT = temperature difference across the thickness of the surface = 
dx = t = thickness normal to the surface = 


2.
- air spacing between two glass panes,

- area of each glass pane,

- thermal conductivity of air,

- temperature difference between the surfaces,

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>



Complete Question
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.
Answer:
The elongation is 
Explanation:
In order to gain a good understanding of this solution let define some terms
True Stress
A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as
.
True Strain
A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as
.
The mathematical relation between stress to strain on the plastic region of deformation is

Where K is a constant
n is known as the strain hardening exponent
This constant K can be obtained as follows

No substituting
from the question we have


Making
the subject from the equation above




From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

Where
is the instantaneous length
is the original length



We can also obtain the elongated length mathematically as follows



Answer: the standard deviation STD of machine B is s (Lb) = 0.4557
Explanation:
from the given data, machine A and machine B produce half of the rods
Lt = 0.5La + 0.5Lb
so
s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)
but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)
so we substitute
s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)
0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)
0.64 = 0.25 + s²(Lb) + 0.4s(Lb)
s²(Lb) + 0.4s(Lb) - 0.39 = 0
s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2
s (Lb) = 0.4557
therefore the standard deviation STD of machine B is s (Lb) = 0.4557