The answer is 20
Your welcome
It's difficult to make out what the force and displacement vectors are supposed to be, so I'll generalize.
Let <em>θ</em> be the angle between the force vector <em>F</em> and the displacement vector <em>r</em>. The work <em>W</em> done by <em>F</em> in the direction of <em>r</em> is
<em>W</em> = <em>F</em> • <em>r</em> cos(<em>θ</em>)
The cosine of the angle between the vectors can be obtained from the dot product identity,
<em>a</em> • <em>b</em> = ||<em>a</em>|| ||<em>b</em>|| cos(<em>θ</em>) ==> cos(<em>θ</em>) = (<em>a</em> • <em>b</em>) / (||<em>a</em>|| ||<em>b</em>||)
so that
<em>W</em> = (<em>F</em> • <em>r</em>)² / (||<em>F</em>|| ||<em>r</em>||)
For instance, if <em>F</em> = 3<em>i</em> + <em>j</em> + <em>k</em> and <em>r</em> = 7<em>i</em> - 7<em>j</em> - <em>k</em> (which is my closest guess to the given vectors' components), then the work done by <em>F</em> along <em>r</em> is
<em>W</em> = ((3<em>i</em> + <em>j</em> + <em>k</em>) • (7<em>i</em> - 7<em>j</em> - <em>k</em>))² / (√(3² + 1² + 1²) √(7² + (-7)² + (-1)²))
==> <em>W</em> ≈ 5.12 J
(assuming <em>F</em> and <em>r</em> are measured in Newtons (N) and meters (m), respectively).
The answer is D)6x2 square units.
A surface area of a cube is a sum of its sides' areas. A side of the cube is a square, and there are total 6 sides of the cube. Also, an area of the side of the cube is the area of the square, which can be expressed as a², where a is the length of the side. Therefore, the surface area (SA) of the cube is:
SA = 6*a² = 6a²
If side length is x, that means that a = x units.
After replacing it in the formula, you will have:
SA = 6a² = 6x² square units
Answer:
X = 58°
Step-by-step explanation:
105° = (2x -11)°
105° + 11° = 2x
116°= 2x
Divide both sides by 2
X = 58°.
Answer:
<h3>Adjacent angles or Supplementary angles I am not sure
tho:)</h3>
Step-by-step explanation:
