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The shear stress at any given point y1 along the height of the cross section is calculated by: where Ic = b·h3/12 is the centroidal moment of inertia of the cross section. The maximum shear stress occurs at the neutral axis of the beam and is calculated by: where A = b·h is the area of the cross section.
Answer:
slope of the equation = -0.006778 and the intercept = 27.11
Explanation:
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The detailed steps and calculation is as shown in the attached file.
Answer:
Explanation:
Force on Q₃ due to charge Q₁
= 9 x 10⁹x 4 x 10⁻⁹ x1 x 10⁻⁹ / 5²
= 1.44 x 10⁻⁹ N
Force due to Q₂ will also be 1.44 x 10⁻⁹ N
component of these forces along x axis
-= 2 x 1.44 x 10⁻⁹ cosθ
= 2.88 x 10⁻⁹ x 4/5
= 2.30x10⁻⁹ N along x axis.
The y-component will calcel out.
b ) In this case , Q₁ will repel and Q₂ will attract.
In this case Q₁ will repel and Q₂ will attract. Component along y - axis will be same as earlier one or 2.30x10⁻⁹ N . Component along x axis will cancel out.
c ) Electric field in case 1 and case 2 will be
= 2.30x10⁻⁹ / 1 x 10⁻⁹
= 2.3 N / C , because field is force per unit charge. The sane field will be in case 2 .
Answer:
Do you need shown work for all of this?