Answer:
w= 62.75 J
Explanation:
Given that
Force vector F= 5 x i + 4 y j
Space vector or displacement vector d= 5.01 i
We know that work (w)
w=∫ F.ds
w= ∫(5 x i + 4 y j) .dx ( only object is moving in x- direction)


![w=\left [\dfrac{5}{2}x^2\right ]_0^{5.01}](https://tex.z-dn.net/?f=w%3D%5Cleft%20%5B%5Cdfrac%7B5%7D%7B2%7Dx%5E2%5Cright%20%5D_0%5E%7B5.01%7D)
w= 2.5 x 5.01² J
w= 62.75 J
Answer: 4,438.96m
Explanation:
(kindly find attachment below)
From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.
By using phythagoras theorem
H² = O² + A²
(5.38)² = (3.20)² + B²
28.944 = 10.24 + B²
B² = 28.944 - 10.24
B² = 18.7044
B = √18.7044
B = 4.439km to meter is 4.439 * 1000 = 4,438. 96m
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The total capacitance is <em>C</em> such that
1/<em>C</em> = 1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)
Solve for <em>C</em> :
<em>C</em> = 1 / (1/(5.0 µF) + 1/(14 µF) + 1/(21 µF)) ≈ 3.1 µF
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