software engineers
hardware engineers
metallurgic engineers
biomechanical engineers
Answer:
12
Explanation:
19-7 lol this is to easy im in middle school
Answer:
axial stress in bar B = 25Mpa.
Deformation of bar A = 0.4mm.
Explanation:
PS: Kindly check the attached picture for the diagram showing the two bars that is to say the bar A and the bar B.
So, we are given the following data or information or parameters which we are going to use in solving this particular question or problem. Here they are;
The cross-sectional areas of Bars A and B = 400 mm2, the modulus of elasticity of bar A and bar B = 200 GPa, applied force = 10kN.
STEP ONE: The first step is to determine or calculate the axial stress in bar B. Therefore,
Axial stress in bar B = 10 × 10³ ÷ 400 × 10⁻⁶ = 25 Mpa.
STEP TWO: The second step here is to determine or calculate the deformation of bar A. Therefore,
The deformation of bar A = 20 × 10³ ×1.5 ÷ 400 × 10⁻⁶ × 200 × 10³ = 0.375 mm.
When the handler has reached the point of balance of the animal, they should stop walking so that they may move only one animal at a time. The point of balance for cattle that are being worked with in limited places such as races and chutes is often found at the animal's shoulder, as seen by the diagrams.
This is further explained below.
<h3>What is a chute?</h3>
Generally, Chutes are channels, planes, or passageways that are either vertical or slanted and through which things may be transported using only gravity.
In conclusion, When the handler has reached the point where the animal's point of balance has been crossed, they should stop walking so that they may move just one animal.
As seen in the illustrations, the point of balance for cattle being worked with in restricted spaces like races and chutes is often located at the animal's shoulder.
Read more about chute
brainly.com/question/10976905
#SPJ1