<span>Astatine same as silicon</span>
To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as
Where,
P = Tensile Force
L= Length
A = Cross sectional Area
E = Young's modulus
PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then
Therefore the elongaton of the rod in a 200mm gage length is 
PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,
Where,
Poission's ratio
= Lateral strain
= Linear strain
Therefore the change in diameter of the rod is 
Answer:
Explanation:
Total distance covered = 25 + 15 = 40 m
Total time = 15 + 8 = 23 s
Average speed = total distance covered / total time
= 40 / 23
= 1.74 m / s
Total displacement = 25 - 15 = 10 m
Total time = 15 + 8 = 23 s
Average velocity = total displacement / total time
= 10 / 23
= .434 m / s to the right .
<h2>Answer:</h2>
<u>Acceleration is </u><u>-10.57 rad/s² </u>
<u>Time is </u><u>6 seconds</u>
<h2>Explanation:</h2><h3>a) </h3>
u=900rpm= 94.248 rad/s
v =300rpm= 31.416 rad/s
s=60 revolutions= 377 rad
v²= u²+ 2as
31.416² = 94.248²+ 2 * a * 377
a = v²-u² / 2s
a= -10.57 rad/s²
<h3>b) </h3>
Using 1st equation of motion
v-u/a = t
Putting the values
t = (31.4 - 94.2)/-10.57
t = 6 seconds
The formula relevant for this is:
h = v0t + 0.5 gt^2
since the rock was dropped, therefore:
h = 0.5 gt^2
we can see that:
h / t^2 = 0.5 g = constant
therefore:
4.9 m / (1 s)^2 = h / (3 s)^2
<span>h = 44.1 m </span>
