What is the net force at the equilibrium point? Derive an equation for the location of the equilibrium point based on the accele
1 answer:
To find a general equilibrium point for a spring based on the hook law, it is possible to start from the following premise:
Hook's law is given by:
Where,
k= Spring Constant
Change in Length
F = Force
When there is a Mass we have two force acting at the System:
W= mg
Where W is the force product of the weigth. Then the force net can be defined as,
But we have a system in equilibrium, so
We find the equilibrium for any location when
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Answer:
The deformation is 0.088289 m
The final height of the monument is 170-0.088289 = 169.911702 m
Explanation:
Thermal coefficient of marble varies between (5.5 - 14.1) ×10⁻⁶/K = α
So, let us take the average value
(5.5+14.1)/2 = 9.8×10⁻⁶ /K
Change in temperature = 35-(-18) = 53 K = ΔT
Original length = 170 m = L
Linear thermal expansion
The deformation is 0.088289 m
The final height of the monument is 170-0.088289 = 169.911702 m (subtraction because of cooling)
Answer:
Δt =
Explanation:
Hi there!
Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.
The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):
v = d/t
Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:
Vr = D/ (2 · Δt1)
For the other half of the trip the expression of velocity will be:
Vw = D/(2 · Δt2)
The total time traveled is the sum of both Δt:
Δt(total) = Δt1 + Δt2
Then, solving the first equation for Δt1:
Vr = D/ (2 · Δt1)
Δt1 = D/(2 · Vr)
In the same way for the second equation:
Δt2 = D/(2 · Vw)
Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)
Δt(total) = D/2 · (1/Vr + 1/Vw)
The time needed by Rick to complete the trip was:
Δt(total) = D/2 · (1/Vr + 1/Vw)
Now let´s calculate the time it took Tim to do the trip:
Tim walks half of the time, then his speed could be expressed as follows:
Vw = 2d1/Δt Where d1 is the traveled distance.
Solving for d1:
Vw · Δt/2 = d1
He then ran half of the time:
Vr = 2d2/Δt
Solving for d2:
Vr · Δt/2 = d2
Since d1 + d2 = D, then:
Vw · Δt/2 + Vr · Δt/2 = D
Solving for Δt:
Δt (Vw/2 + Vr/2) = D
Δt = D / (Vw/2 + Vr/2)
Δt = D/ ((Vw + Vr)/2)
Δt = 2D / (Vw + Vr)
The time needed by Tim to complete the trip was:
Δt = 2D / (Vw + Vr)
Let´s find the diference between the time done by Tim and the one done by Rick:
Δt(tim) - Δt(rick)
2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))
= Δt
Let´s check the result. If Vr = Vw:
Δt = 2D/2Vr - D/2Vr - D/2Vr
Δt = D/Vr - D/Vr = 0
This makes sense because if both move with the same velocity all the time both will do the trip in the same time.