They compress or expand depending on amount of pressure or depending on the temperature
Answer:
Explanation:
Given that:
width b=100mm
depth h=150 mm
length L=2 m =200mm
point load P =500 N
Calculate moment of inertia

Point C is subjected to bending moment
Calculate the bending moment of point C
M = P x 1.5
= 500 x 1.5
= 750 N.m
M = 750 × 10³ N.mm
Calculate bending stress at point C

Calculate the first moment of area below point C

Now calculate shear stress at point C


Calculate the principal stress at point C
![\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa](https://tex.z-dn.net/?f=%5Csigma_%7B1%2C2%7D%3D%5Cfrac%7B%5Csigma_x%2B%5Csigma_y%7D%7B2%7D%20%5Cpm%5Csqrt%7B%28%5Cfrac%7B%5Csigma_x-%5Csigma_y%7D%7B2%7D%20%29%20%2B%20%28%5Ctau%29%5E2%7D%20%5C%5C%5C%5C%3D%5Cfrac%7B666.67%2B0%7D%7B2%7D%20%5Cpm%5Csqrt%7B%28%5Cfrac%7B666.67-0%7D%7B2%7D%20%29%5E2%20%5Cpm%2844.44%29%5E2%7D%20%5C%20%5B%20%5Csigma_y%3D0%5D%5C%5C%5C%5C%3D333.33%5Cpm336.28%5C%5C%5C%5C%20%5Csigma_1%3D333.33%2B336.28%5C%5C%3D669.61KPa%5C%5C%5C%5C%5Csigma_2%3D333.33-336.28%5C%5C%3D-2.95KPa)
Calculate the maximum shear stress at piont C

Answer:
a) 16m/s b) 192m
Explanation:
v1=32m/s a=-2m/s^2 t=8s v2=? d=??
a) I will use this equation v2= v1 + a*t
v2= 32m/s + -2m/s^2 * 8s
v2= 32m/s + -16m/s
v2= 16m/s
b) v2^2=v1^2 + 2ad
rearranging
v2^2-v1^2=2ad
v2^2-v1^2/2= a d
v2^2-v1^2/2a=d
16m/s^2 - 32m/s^2/ 2 x-2m/s^2 =d
d=192m
By the Newtons Second Law:
F = ma
F = (47 kg)(15 m/s²)
<h2>F = 705 N</h2>