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).
Step-by-step explanation:
Here is the answer to your question.
Answer:
2500 tiles
Step-by-step explanation:
convert the path to the same units, one m = 100 cm
path= 500 cm x 300 cm
path area = 150,000 cm^2
each tile = 10x6 = 60 cm^2
150,000/60 = 2500 tiles
Answer:
m∠BEF = 65.3°
Step-by-step explanation:
Given:
m∠DEB = 27.2,
m∠DEF = 92.5
Required:
m∠BEF
SOLUTION:
Since B is the interior of ∠DEF, it means ∠DEB and ∠BEF are adjacent angles that make up ∠DEF. And they share the same side, BE.
Therefore:
m∠BEF + m∠DEB = m∠DEF (angle addition postulate)
m∠BEF + 27.2 = 92.5
Subtract 27.2 from each side
m∠BEF + 27.2 - 27.2 = 92.5 - 27.2
m∠BEF = 92.5 - 27.2
m∠BEF = 65.3°
