This question is loaded with misleading, marginally meaningful phrases.
A 12V battery produces more current than a 1.5V battery produces,
through the same device, because the potential difference between
the terminals of the 12V battery is 8 times the potential difference
between the terminals of the 1.5V battery.
Of the choices offered, choice 'C' is the closest. But it should say
"electrical potential", NOT "potential energy".
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)
Answer:
a=
Explanation:
The net force, 
of the box is expressed as a product of acceleration and mass hence
where m is mass and a is acceleration
Making a the subject, a= 
From the attached sketch,
∑ 
where 
is frictional force and 
is horizontal angle
Substituting ∑ as in the equation where we made a the subject
a= 
Since we’re given the value of F as 240N, as 41.5N, as 
and mass m as 30kg
a= 
Answer:
F = 118 N
Explanation:
Assume Ann and Bob lift at their respective ends of the table
Sum moments about Bob's position to zero.
Let F be Ann's upward force
F[2.25] - 18.5(9.80)[2.25 / 2] - 8.33(9.80)[0.750] = 0
F = 117.86133333...
