Answer:
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Answer:
W = -0.480 J
Explanation:
given,
q₁ = 4 μC
q₂ = -4.10 μC


b = 0.381
k = 8.99 × 10⁹ Nm²/C²

![W = [-147.436\times (5.88-2.62)\times 10^{-3}]J](https://tex.z-dn.net/?f=W%20%3D%20%5B-147.436%5Ctimes%20%285.88-2.62%29%5Ctimes%2010%5E%7B-3%7D%5DJ)
W = -0.480 J
Work done by the electric force W = -0.480 J
<span>The formulas are,
v1d1² = v2d2² ........ (1)
h = (v2²-v1²)/2g ...... (2)
Given that,
v1 = 1.71 m/s
we assume that the stream has decreased by a factor
d2 =0.805d1
then,
v1d1² = v2 (0.805d1)²
cancelled both side d1² then we get,
v1 = v2 (0.805)²
v1 = v2 (0.648025)
Sub v1 = 1.71,
1.71 = v2 (0.648025)
v2 = 1.71/0.648025
v2 = 2.638787083831642
v2 = 2.64 m/s
The vertical distance formula,
h = (v2²-v1²)/2g
We know that value of gravity constant is 9.8 m/s²
h = {(2.64)² - (1.71)²)/2(9.8)
h = {(6.9696) - (2.9241)}/19.6
h = (4.0455)/19.6
h = 0.2064030612244898
h = 0.21 cm
Therefore, the vertical distance h = 0.21 cm.</span>
B because willingness shows confidence in your answer
Answer:
Option B: change the objective lens
Explanation:
The revolving nosepiece is one of the parts of a microscope. Its responsibility is to hold the objective lenses.