The force in the rod when the temperature is 150 °F is 718.72 pounds-force.
<h3>How to determine the resulting the resulting force due to mechanical and thermal deformation</h3>
Let suppose that rod experiments a <em>quasi-static</em> deformation and that both springs have a <em>linear</em> behavior, that is, force (
), in pounds-force, is directly proportional to deformation. Then, the elongation of the rod due to <em>temperature</em> increase creates a <em>spring</em> deformation additional to that associated with <em>mechanical</em> contact.
Given simmetry considerations, we derive an expression for the <em>spring</em> force (
), in pounds-force, as a sum of mechanical and thermal effects by principle of superposition:
(1)
Where:
- Spring constant, in pounds-force per inch.
- Spring deformation, in inches.
- Rod elongation, in inches.
The <em>rod</em> elongation is described by the following <em>thermal</em> dilatation formula:
(2)
Where:
- Coefficient of linear expansion, in
.
- Initial length of the rod, in inches.
- Initial temperature, in degrees Fahrenheit.
- Final temperature, in degrees Fahrenheit.
If we know
,
,
,
,
and
, then the force in the rod at final temperature is:
![F = \left(1000\,\frac{lb}{in} \right)\cdot \left[0.7\,in + 0.5\cdot\left(6.5\times 10^{-6}\,\frac{1}{^{\circ}F} \right)\cdot (48\,in)\cdot (150\,^{\circ}F-30\,^{\circ}F)\right]](https://tex.z-dn.net/?f=F%20%3D%20%5Cleft%281000%5C%2C%5Cfrac%7Blb%7D%7Bin%7D%20%5Cright%29%5Ccdot%20%5Cleft%5B0.7%5C%2Cin%20%2B%200.5%5Ccdot%5Cleft%286.5%5Ctimes%2010%5E%7B-6%7D%5C%2C%5Cfrac%7B1%7D%7B%5E%7B%5Ccirc%7DF%7D%20%5Cright%29%5Ccdot%20%2848%5C%2Cin%29%5Ccdot%20%28150%5C%2C%5E%7B%5Ccirc%7DF-30%5C%2C%5E%7B%5Ccirc%7DF%29%5Cright%5D)

The force in the rod when the temperature is 150 °F is 718.72 pounds-force. 
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LINES AND ANGLES
After observing we get as ..Going horizontal manner we get
1) Alternate Exterior Angles
2) None of these
3) Corresponding Angles
4)Alternate interior Angles
5) Corresponding Angles
Answer:
6 square units
Step-by-step explanation:
In oder to do this, you need graph paper to make it easier and then mark your graph paper with the quadrants.
I got 3 unites for the height and 4 units for the base
4 x 3 = 12 square units
however the formula for a triangle is Ab/2
so divide 12/2 = 6
So its 6 square units
6 2/3 divided equally into containers holding 1 1/3 would be 1 1/3 + 1 1/3 + 1 1/3 + 1 1/3+ 1 1/3= 6 2/3. That would fill five containers with none leftover.