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 = -5/6 m/s²
Explanation:
v² = u² + 2as
a = (v² - u²) / 2s
a = (0² - 5²) / 2(15)
a = -5/6 m/s²
This assumes the initial velocity was in the positive direction.
Answer
A) Positively charged two insulating rod are brought closed to an object they repel each other. It means the Object is positively charged. Because similar charge repel each other.
The correct answer is Option A.
B) we know force between to charges is calculated using Formula
.......(1)
form the given condition in the question
from equation (1)
hence, the correct answer is Option C.
Answer:
6kg m/s
Explanation:
Momentum = p=mxv
Momentum = mass x velocity
4kg x 1.5m= 6 kg m/s
