The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μr=0.002. Suppose a 180
1 answer:
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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
Answer:
The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³
Explanation:
Given that,
Volume of seawater = 90 % of Volume of iceberg
Volume of visible surface = 10 %
Density of seawater = 1025 kg/m³
We need to calculate the density of the iceberg
Using equilibrium condition
Put the value into the formula
We need to calculate the density of the iceberg
Using equilibrium condition
Put the value into the formula
Hence, The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³