Answer:
too little body fat is a health risk
Explanation:
IF you do not have enough body fat you can have bad things happen to you
The bunch of stuff in the jar is a <em>mixture. </em> The different kinds of pieces aren't bound to each other, and they can be easily separated.
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 \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7B4v%7D%7B%203%20A%7D%20%3D%20%5Cfrac%7B4%20%2875%20%5C%20lbf%20%2B%2025%20%5C%20lbf%29%7D%7B%20%5Cfrac%7B3%20%28%20lni%29%5E2%7D%7B4%7D%7D)
![= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B4%20%28100%20%5C%20lbf%29%7D%7B%20%5Cfrac%7B3%20%28%20lni%29%5E2%7D%7B4%7D%7D%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7B400%20%5C%20lbf%29%7D%7B%20%5Cfrac%7B3%20%28%20lni%29%5E2%7D%7B4%7D%7D%20%5C%5C%5C%5C%3D%200.17%20%5C%20ksi)
![= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in](https://tex.z-dn.net/?f=%3D%20R%20%5Ctimes%20200%20%5C%20in%20-%20P%20%5Ctimes%20100%20%5C%20in%20%3D%2012500%20%5C%20lbf%20%5Ctimes%5C%20%20in)
![\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)
To solve the problem, we
must know the heat capacity of ice and water.
For Cp = 2090 J/kg C
H = mCpT
H = (10 kg) ( 2090 J/ Kg C)
( -23 C)
H = - 480700 J
For water Cp = 4180 j/kg C
H = (100 kg) ( 4180 J/kg C)
( 60 C)
<span>H = 2508000 J</span>