(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
<h3>
Work done in the spring</h3>
The work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
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Answer:
0.02
Explanation:
coefficient of kinetic friction = μ
force of friction = Ff
Normal Force = FN, but
FN = -W
Ff = -μFN
so μ = Ff/FN
= 4N/200N
= 0.02.
Answer:
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Explanation:
Answer:
18 Ω
Explanation:
As K and F are at the same voltage, we can redraw the diagram as in figure 2
Series resistances add directly, so we get figure 3
Adding parallel resistances gets us to figure 4
Now we can move two 6Ω resistances for clarification in figure 5
As the voltage between C and J will be identically split between D and H, there will be no voltage drop across the middle 6Ω resister and no current through it, identical to an infinite resistance, so that 6Ω can be eliminated as in figure 6
Add series resistances to get to figure 7
Add parallel resistances to get to figure 8
Add series resistances to get to figure 9
F = ma, where m = mass in kg, a = acceleration in m/s², F = Force in Newton
F = 1 * 2
F = 2 N
Force needed is 2 Newtons.
