A spring with a spring constant
1 answer:
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To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:
Where,
P = Pressure
V = Volume
T = Temperature
n = amount of substance
R = Ideal gas constant
We start by calculating the volume of inhaled O_2 for it:
Our values are given as
P = 1atm
T=293K
Using the equation to find n, we have:
Number of molecules would be found through Avogadro number, then
Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²