**Problem 1**

Calculate how much of a “unit” of sunlight the panel absorbs at three different times along the sun’s path.

Panel Vectorp=⟨ 7, −3⟩

u1=⟨−0.38, −0.92⟩

u2=⟨−0.5, − 0.87⟩

u3=⟨−0.7, −0.7⟩

**For u1**= ⟨−0.38, −0.92⟩

u1⋅p= (−0.38) ⋅7+(−0.92) ⋅ (−3) u1⋅p= (−2.66) + (2.76) = 0.10

**u2**=⟨−0.5, −0.87

p= (−0.5) ⋅7+ (−0.87) ⋅ (−3)

p= (−3.5) + (2.61) =−0.89

=⟨−0.7, − 0.7⟩\

= (−0.7) ⋅7+(−0.7) ⋅ (−3) (−4.9)+(2.1)=−2.80

**Summary of Dot Products**

- u1⋅p=0.10
- u2⋅p=−0.89
- u3⋅p=−2.80

vector u1, which is <-0.38, -0.92>, has the highest dot product, at -0.38. This implies that a rotation aligning the panel with this vector, corresponding to a 25% rotation, will yield the most sunlight reception.

**Justification for Maximum Sunlight Absorption**

The amount of sunlight the panel absorbs is related to the cosine of the angle between the sun ray vector and the normal vector “u.” Cosine (theta) = (Sun Ray Vector) • (Normal Vector “u”) / ||Sun Ray Vector|| * ||Normal Vector “u”||.

When the sun ray strikes perpendicularly (like Sun Ray 1), the angle (theta) is close to 0 degrees, and cosine (theta) approaches 1. This signifies maximum absorption (close to one unit).

- As the angle increases (like Sun Ray 2), cosine (theta) decreases, resulting in less sunlight absorption (less than one unit).
- When the sun ray grazes the panel (like Sun Ray 3), the angle approaches 90 degrees, and cosine (theta) approaches 0. This indicates minimal sunlight absorption (minimal).

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**Task 2-Problem 2**

Distance between (1, 1, 9) and (x, y, 6):

d1 = sqrt ((1 – x)^2 + (1 – y)^2 + (9 – 6)^2)

Distance between (-1, 1, 8) and (x, y, 6):

d2 = sqrt((1 + x)^2 + (1 – y)^2 + (8 – 6)^2)

Distance between (1, -1, 10) and (x, y, 6):

d3 = sqrt((1 – x)^2 + (-1 – y)^2 + (10 – 6)^2)

Since we want all cables to have the same length, we set these distances equal to each other:

d1 = d2

d1 = d3

Setting Distances Equal

- d1 = d2: This expands to (1 – x)^2 + (1 – y)^2 = (1 + x)^2 + (1 – y)^2. The (1 – y)^2 terms cancel out.
- d1 = d3: This expands to (1 – x)^2 + (1 – y)^2 = (1 – x)^2 + (-1 – y)^2. Again, (1 – x)^2 terms cancel out.

(x, y) = (0, 0).

**Stoplight Location**

Knowing (x, y) = (0, 0) and the stoplight’s z-coordinate is fixed at 6, the final stoplight location that minimizes cable tension is:**(0, 0, 6)**

**Task 3**

Force exerted by her back at 100% rotation (F1): 539N

For the “Improper Technique” at 30% rotation

The force exerted by her back at 30% rotation(F2): 294N

To calculate the work done during the lifting motion from 0% to 100% rotation in the “Improper Technique,” we need to use the work formula: Work = Force × Distance

Given

|F1|=539N

|r1|=0.61m

|F2|=294N

|r2|=0.72m

Work done during the lifting motion:

(a) for the first segment (0% to 30% rotation)

Work1 =

F2×Distance1

W1W1W1=F2×|r2|=294N×0.72m=211.68J

(b) for the second segment (30% to 100% rotation)

Work2 =

F2×Distance2

W2W2W2=F1×|r1|=539N×0.61m=328.79J

Total work done,

Total work =W1+W2

Total work =211.68+328.79

Total work =540.47 Joules

The total work done during the lifting motion, Total work =540.47 Joules. Therefore, the total work done during the lifting motion in the “Improper Technique” is 540.47 Joules.