The first one is A and the second one would be C
Answer:
2.17 Mpa
Explanation:
The location of neutral axis from the top will be

Moment of inertia from neutral axis will be given by 
Therefore, moment of inertia will be
![\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}](https://tex.z-dn.net/?f=%5Cfrac%20%7B240%5Ctimes%2025%5E%7B3%7D%7D%7B12%7D%2B%28240%5Ctimes%2025%29%5Ctimes%20%2856.25-25%2F2%29%5E%7B2%7D%2B2%5Ctimes%20%5B%5Cfrac%20%7B20%5Ctimes%20150%5E%7B3%7D%7D%7B12%7D%2B%2820%5Ctimes%20150%29%5Ctimes%20%28%2825%2B150%2F2%29-56.25%29%5E%7B2%7D%5D%3D34.5313%5Ctimes%2010%5E%7B6%7D%20mm%5E%7B4%7D%7D)
Bending stress at top= 
Bending stress at bottom=
Mpa
Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa
Answer:
<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>
Explanation:
The image is shown below.
The force on the particle with charge q due to each charge Q = 
we designate this force as N
Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.
Resolving the forces on the particle, we have
for the x-component
= N cosine 60° + (-N cosine 60°) = 0
for the y-component
= -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N
The above indicates that there is no resultant force in the x-axis, since it is equal to zero (
= 0).
The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.
<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>
Two identical balls collide<span> head on. The </span>initial velocity<span> of </span>one<span> is 0.75 </span>m/s<span> east, while that of the </span>other one<span> is 0.43 </span>m/s west<span>.</span>