Answer:
x = 25 / μ [ ft]
Explanation:
To solve this exercise we can use Newton's second law.
Let's set a reference system where the x axis is parallel to the road
Y axis
N_B + N_A - W_van - W_load = 0
N_B + N_A = W_van + W_load
X axis
fr = ma
a = fr / m
the total mass is
m = (W_van + W_load) / g
the friction force has the expression
fr = μ N_{total}
fr = μy (W_van + W_load)
we substitute
a = μ (W_van + W_load)
a = μ g
taking the acceleration let's use the kinematic relations where the final velocity is zero
v² = v₀² - 2 a x
0 = v₀² -2a x
x =
x =
x =
x = 25 / μ [ ft]
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Explanation: <em>A mixture in which its constituents are distributed uniformly is called homogeneous mixture, such as salt in water. A mixture in which its constituents are not distributed uniformly is called heterogeneous mixture, such as sand in water.</em>
Answer:
V = 6.36 m³
Explanation:
For this exercise we will use fluid mechanics relations, starting with the continuity equation.
Let's write the flow equation
Q = v₁ A₁
The area of a circle is
A = π r²
Radius is half the diameter
A = π/4 d²
Q = v₁ π/4 d₁²
Q = π/ 4 15 0.03 2
Q = 0.0106 m3 / s
The volume of water in t = 10 min = 10 60 = 600 s
Q = V / t
V = Q t
V = 0.0106 600
V = 6.36 m³
Speed = Distance/Time. So you are required to know A. Distance and period of time