high velocity is the answer as baby oil is dense
Answer:
a) W = 6.75 J and b) v = 3.87 m / s
Explanation:
a) In the problem the force is nonlinear and they ask us for work, so we must use it's definition
W = ∫ F. dx
Bold indicates vectors. In a spring the force is applied in the direction of movement, whereby the scalar product is reduced to the ordinary product
W = ∫ F dx
We replace and integrate
W = ∫ (-60 x - 18 x²) dx
W = -60 x²/2 -18 x³/3
Let's evaluate between the integration limits, lower W = 0 for x = -0.50 m, to the upper limit W = W for x = 0 m
W = -30 [0- (-0.50) 2] -6 [0 - (- 0.50) 3]
W = 7.5 - 0.75
W = 6.75 J
b) Work is equal to the variation of kinetic energy
W = ΔK
W = ΔK = ½ m v² -0
v =√ 2W/m
v = √(2 6.75/ 0.90)
v = 3.87 m / s
We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is
X=0m/s
From the question we are told that
mosquito flying 2 m/s
against a 2 m/s headwind
Generally
The speed over the ground is the Flight Speed minus resistance speed
Generally the equation for the speed over the ground is mathematically given as
X=Flight Speed-resistance speed
Therefore
X=2-2
X=0m/s
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