What is the shortest possible time in which a bacterium could travel a distance of 8.4cm across a Petri dish at a constant speed
1 answer:
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Answer:
The answer is 1.87nm/s.
Explanation:
The water loss must be replaced by
of sap. 110g of sap corresponds to a volume of
thus rate of sap replacement is
The volume of sap in the vessel of length is
,
where is the cross sectional area of the vessel.
For 2000 such vessels, the volume is
taking the derivative of both sides we get:
on the left-hand-side is the velocity
of the sap, and on right-hand-side
; therefore,
and since the cross-sectional area is
;
therefore,
solving for we get:
which is the upward speed of the sap in each vessel.
Answer:
a) α = 1.875
b) t = 8 s
Explanation:
Given:
ω1 = 0
ω2 = 15
theta (angular displacement) = 60 rad
*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*
a) α = ?
(ω2)^2 = (ω1)^2 + 2α(theta)
=
+ 2(α)(60)
225 = 120α
α = 1.875
b)
α = (ω2-ω1)/t
t = (ω2-ω1)/α = (15-0)/1.875 = 8
t = 8 s