Answer:
The answer is "effective stress at point B is 7382 ksi
"
Explanation:
Calculating the value of Compressive Axial Stress:
![\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\](https://tex.z-dn.net/?f=%5Cto%20%5Csigma%20y%20%20%3D%5Cfrac%7BF%7D%7BA%7D%20%3D%20%5Cfrac%7B4%20F%7D%7B%28%20p%20d%20%5E2%20%29%7D%20%3D%20%5Cfrac%7B%284%20x%20%28%20-%2040000%20%5C%20lbf%29%29%7D%7B%5B%20p%20%5Ctimes%20%281%20%5C%20in%29%5E2%20%5D%7D%20%3D%20-%2050.9%20%5C%20ksi%20%5C%5C)
Calculating Shear Transverse:
![\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\](https://tex.z-dn.net/?f=%5Cto%20%5Csigma%27%20%3D%5B%20s%20y%5E2%20%2B3%28%20t%20%5Ctimes%20y%5E2%20%2B%20t%20yz%5E2%20%29%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C)
![= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi](https://tex.z-dn.net/?f=%3D%20%5B%20%28-50.9%29%5E2%20%2B3%28%2863.7%29%5E2%20%2B%280.17%29%5E2%20%29%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%5B2590.81%2B%203%284057.69%29%2B0.0289%5D%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%5B2590.81%2B12%2C173.07%2B0.0289%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D14763.9089%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%20%3D%207381.95445%20%5C%20ksi%5C%5C%5C%5C%20%3D%207382%20%5C%20ksi)
<u>Explanation</u>:
According to Newton's third law that each force has an equal and opposite reaction force in this case both of the jars will exert the same force an each other
. The force is given by
Where, F = force, 
, 
, 
and Distance(r) = 0.42 m.
Substitute the values in the formula.
The answer is 0k because 143c equals nothing
Answer:
a= 17.69 m/s^2
Explanation:
Step one:
given data
A car accelerates uniformly from rest to 23 m/s
u= 0m/s
v= 23m/s
distance= 30m
Step two:
We know that
acceleration= velocity/time
also,
velocity= distance/time
23= 30/t
t= 30/23
t= 1.30 seconds
hence
acceleration= 23/1.30
accelaration= 17.69 m/s^2
When driving on roads that may be slippery, do not make any sudden changes in speed or direction. Option D is correct.
<h3 /><h3>What is a slippery surface?</h3>
The slick road sign serves as a warning. When the road is wet or ice, drivers should use extra caution and reduce their speed. When the weather is bad, avoid making any rapid changes in direction.
When driving on roads that may be slippery, do not make any sudden changes in speed or direction. It may cause accident because the vehicle can lose their balance.
Hence, option D is correct.
To learn more about the slippery surface, refer to the link;
brainly.com/question/1953680
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