434543645445 x 536436345363 please help my college teacher want's me to do this hurry asap.

Mathematics

2 answers:

gizmo_the_mogwai [7]3 years ago

8 0

Answer:

from what google says the answer should be 2.3310501e+23?

Step-by-step explanation:

I'm pretty sure thats correct

Luba_88 [7]3 years ago

6 0

Answer:

2.3310501e+23

Step-by-step explanation:

i dunno i googled it

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The rod is made of A-36 steel and has a diameter of 0.22 in . If the rod is 4 ft long when the springs are compressed 0.7 in . a

ziro4ka [17]

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 quasi-static deformation and that both springs have a linear behavior, that is, force ( $F$ ), in pounds-force, is directly proportional to deformation. Then, the elongation of the rod due to temperature increase creates a spring deformation additional to that associated with mechanical contact.

Given simmetry considerations, we derive an expression for the spring force ( $F$ ), in pounds-force, as a sum of mechanical and thermal effects by principle of superposition:

$F = k\cdot (\Delta x + 0.5\cdot \Delta l)$ (1)

Where:

$k$ - Spring constant, in pounds-force per inch.
$\Delta x$ - Spring deformation, in inches.
$\Delta l$ - Rod elongation, in inches.

The rod elongation is described by the following thermal dilatation formula:

$\Delta l = \alpha \cdot L_{o}\cdot (T_{f}-T_{o})$ (2)

Where:

$\alpha$ - Coefficient of linear expansion, in $\frac{1}{^{\circ}F}$ .
$L_{o}$ - Initial length of the rod, in inches.
$T_{o}$ - Initial temperature, in degrees Fahrenheit.
$T_{f}$ - Final temperature, in degrees Fahrenheit.

If we know $k = 1000\,\frac{lb}{in}$ , $\Delta x = 0.7\,in$ , $\alpha = 6.5\times 10^{-6}\,\frac{1}{^{\circ}F}$ , $L_{o} = 48\,in$ , $T_{o} = 30\,^{\circ}F$ and $T_{f} = 150\,^{\circ}F$ , then the force in the rod at final temperature is:

$F = \left(1000\,\frac{lb}{in} \right)\cdot \left[0.7\,in + 0.5\cdot\left(6.5\times 10^{-6}\,\frac{1}{^{\circ}F} \right)\cdot (48\,in)\cdot (150\,^{\circ}F-30\,^{\circ}F)\right]$