Dos automóviles que marchan en el mismo sentido, se encuentran a una distancia de 126 Km. Si el más lento va a 42 Km/h, calcular
1 answer:
Answer:
<em>La velocidad del más rápido deberá ser de 63 Km/h</em>
Explanation:
<u>Movimiento Rectilíneo Uniforme (MRU)</u>
Un cuerpo se dice que tiene MRU cuando recorre iguales distancias en tiempos iguales y en línea recta.
Tenemos dos automóviles viajando en el mismo sentido. Uno, el más lento va por delante del otro que le ancanzará en t=6 horas una vez que supere la ventaja que le lleva el otro.
La velocidad se puede calcular con la fórmula:
Donde x es la distancia y t el tiempo que tarda en recorrerla.
El vehículo más rápido alcanzará al otro cuando logre superar los x=126 Km que le lleva. Si esto lo hace en t=6 horas, entonces la velocidad adicional que deberá desarrollar es:
Esa velocidad se suma a la del primer vehículo y tendremos la velocidad necesaria: 42 Km/h + 21 Km/h = 63 Km/h
La velocidad del más rápido deberá ser de 63 Km/h
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Answer:
a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324
Explanation:
For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces
let's use subscript 1 for the ladder and 2 for the firefighter
∑ τ = 0
-W₁ x₁ - W₂ x₂ + N₁ y = 0
N₁ = (1)
the center of mass of the ladder is at its geometric center,
d = L / 2 = 15/2 = 7.5 m
cos 60 = x₁ / d₁
x₁ = d₁ cos 60
x₁ = 7.5 cos 60
x₁ = 3.75 m
for the firefighter d₂ = 4 m
cos 60 = x₂ / d₂
x₂ = d₂ cos 60
x₂ = 4 cos 60 = 2 m
for the fulcrum d₃ = 15 m
sin 60 = y / d₃
y = d₃ sin 60
y = 15 sin 60
y = 13 m
we look for the Normal by substituting in equation 1
N₂ =
N₂ = 267.3 N
now let's use the translational equilibrium relations
X axis
F₁ - N₂ = 0
F₁ = N₂
F₁ = 267.3 N
Axis y
N₁ - W₁ -W₂ = 0
N₁ = W₁ + W₂
N₁ = 500 + 800
N₁ = 1300 N
b) for this case change the firefighter's distance d₂ = 9 m
x₂ = 9 cos 60
x₂ = 4.5 m
we substitute in 1
N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}
N₂ = 421.15 N
of the translational equilibrium equation on the x-axis
fr = F₁ = N₂
fr = 421.15 N
friction force has the expression
fr = μ N
in this case the reaction of the Earth to the support of the ladder is N1 = 1300N
μ = fr / N₁
μ = 421.15 / 1300
μ = 0.324
Answer:
The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.
Determine Fx."
Explanation:
We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.
torque=cross product of force and position . mathematically this can be express as
Where
and the position vector
using the determinant method to expand the cross product in order to determine the torque we have
by expanding we arrive at
since we have determine the vector value of the toque, we now compare with the torque value given in the question
if we directly compare the j coordinate we have