A large truck and a compct car are both driving down the
1 answer:
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Answer:
Q = 165.95 cm³ / s, 1) v = , 2) v = 2.05 m / s,
3) d₂ = 1.014 cm
Explanation:
This is a fluid mechanics exercise
1) the continuity equation is
Q = v A
where Q is the flow rate, A is area and v is the velocity
the area of a circle is
A = π r²
radius and diameter are related
r = d / 2
substituting
A = π d²/4
Q = π/4 v d²
let's reduce the magnitudes
v = 0.55 m / s = 55 cm / s
let's calculate
Q = π/4 55 1.96²
Q = 165.95 cm³ / s
If we focus on a water particle and apply the zimematics equations
v² = v₀² + 2 g y
where the initial velocity is v₀ = 0.55 m / s
v =
v =
2) ask to calculate the velocity for y = 0.2 m
v =
v = 2.05 m / s
3) We write the continuous equation for this point 2
Q = v₂ A₂
A₂ = Q / v₂
let us reduce to the same units of the SI system
Q = 165.95 cm³ s (1 m / 10² cm) ³ = 165.95 10⁻⁶ m³ / s
A₂ = 165.95 10⁻⁶ / 2.05
A₂ = 80,759 10⁻⁶ m²
area is
A₂ = π/4 d₂²
d₂ =
d₂ =
d₂ = 10.14 10⁻³ m
d₂ = 1.014 cm
Answer:
x = 240 m
Explanation:
This is a kinematics exercise
Let's fix our frame of reference on car A
x = x₀ₐ+ v₀ₐ t + ½ aₐ t²
the initial position of car a is zero
x = 0 + v₀ₐ t + ½ 0.8 t²
for car B
x = x_{ob} + v_{ob} t - ½ a_b t²
car B's starting position is 30 m
x = 30 + v_{ob} t - ½ 0.4 t²
at the point where they meet, the position of the two vehicles is the same
0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²
let's reduce the speeds to the SI system
v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s
v_{ob} = 23.4 km / h = 6.5 m / s
4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²
0.2 t² - 2.5 t - 30 = 0
t² - 12.5 t - 150 = 0
we solve the quadratic equation
t =
t =
t₁ = 20 s
t₂ = -7.5 s
time must be a positive quantity so the correct result is t = 20 s
let's look for the distance
x = 4 t + ½ 0.8 t²
x = 4 20 + ½ 0.8 20²
x = 240 m