Q should be positioned 4.8 miles east of point B. Explanation: From the diagram, we can define the construction cost as a function of angle θ (as illustrated). The underwater pipe length (marked in blue) equals 6/cos θ, while the land pipe length (marked in brown) is (8 - 6*tan θ). The total construction cost formula is: Construction Cost = (6/cos θ)(6000) + (8 - 6*tan θ)(3750). This formula is represented in terms of θ, which can vary from 0 degrees to 53.13 degrees according to the diagram. To find the angle θ that minimizes the construction cost, we differentiate the Construction Cost function with respect to θ and set it to zero. The derivative yields: -4500(5*sec θ – 8*tan θ)(sec θ) = 0, leading to θ = 38.68 degrees. By substituting θ, we can determine the distance of Q from B, which equals 6*tan θ. This calculates to a distance of 4.8 miles.
<span>We will apply the momentum-impulse theorem here. The total momentum along the x-direction is defined as p_(f) = p_(1) + p_(2) + p_(3) = 0.
Therefore, p_(1x) = m1v1 = 0.2 * 2 = 0.4. Additionally, p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3, where v3 represents the unknown speed and m3 signifies the mass of the third object, which has an unspecified velocity.
In the same way, for the particle of 235g, the y-component of the total momentum is described with p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
Thus, p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3, where m3 is the mass of the third piece.
Consequently, p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; yielding v3 = 0.4/-0.1 = - 4.
Similarly, p_(fy) = 0.3525 + 0.1v3; thus v3 = - 0.3525/0.1 = -3.525.
Therefore, the x-component of the speed of the third piece is v_3x = -4 and the y-component is v_3y = 3.525.
The overall speed is calculated as follows: resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
