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Maru [420]
3 years ago
11

Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se

mueve a 5.00 m/s. Sue aplica los frenos, pero solo puede acelerar a -2.00 m/s2 porque el camino esta húmedo. ¿Habrá una colisión? Establezca como llega a su respuesta. Si es si, determine cuán lejos en el túnel y en qué tiempo ocurre la colisión. Si es no, determine la distancia de acercamiento mas próxima entre el automóvil de Sue y la camioneta.
Physics
1 answer:
Romashka [77]3 years ago
0 0

Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}

Camioneta lenta

x_{C} = x_{C,o} +v_{C}\cdot t

Donde:

x_{S,o}, x_{C,o} - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

v_{S,o} - Velocidad inicial de Speedy Sue, medida en metros por segundo.

v_{S}, v_{C} - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

a_{S} - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

t - Tiempo, medido en segundos.

Si conocemos que x_{S} = x_{C}, x_{S,o} = 0\,m, x_{C,o} = 155\,m, v_{S,o} = 30\,\frac{m}{s}, v_{C} =-5\,\frac{m}{s} y a_{S} = -2\,\frac{m}{s^{2}}, encontramos la siguiente función cuadrática:

155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}

-t^{2}+35\cdot t-155 = 0 (Ec. 1)

Las raíces de esta función son:

t_{1}\approx 29.798\,s, t_{2} \approx 5.201\,s

La colisión ocurriría en la raíz positiva más pequeña, es decir:

t \approx 5.201\,s

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: (v_{C,o} = -5\,\frac{m}{s},  x_{C,o} = 155\,m, t \approx 5.201\,s)

x_{C} = 155\,m +\left(-5\,\frac{m}{s}\right)\cdot (5.201\,s)

x_{C} = 128.995\,m

En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

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I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:

When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.

1. For the maximum height, the equation is
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For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
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The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.

2. For this, you equate the y values of both balls:

y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
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Answer:A:The track pushes back on Clinton's shoe with the same force.

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Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi
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Answer:

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Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

     cos ( 360-30) = cos (-30) = x₁ / d

     sin (-30) = y₁ / d

     x₁ = d cos (-30)

     y₁ = d sin (-30)

     x₁ = 1.2 cos (-30) = 1,039 miles

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second shift. d = 2.0 miles to 20º West of South

       cos (270-20) = x₂ / d

       cos (250) = y₂ / d

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       y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

       cos (180 + 30) = x₃ / d

       sin (210) = y₃ / d

       x₃ = 1.6 cos 210 = -1.3856 miles

       y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

       cos (90 + 15) = x₄ / d

       sin (105) = y₄ / d

       x₄ = 2.6 cos 105 = -0.6729 miles

       y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis  (West-East direction)

       X = x₁ + x₂ + x₃ + x₄

       X = 1.039 -0.684 - 1.3846 - 0.6729

       X = -1.7025 miles

   

       Y = y₁ + y₂ + y₃ + y₄

       Y = -0.6 -1.879 -0.8 +2.511

       Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

      D = √ (X² + Y²)

      D = √ (1.7025² + 0.768²)

      D = 1.8677 miles

let's use trigonometry to find the direction

       tan θ = Y / X

       θ = tan⁻¹ Y / x

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       θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

         θ = 24.28º at South of West

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Explanation:

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Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F  = Force, q = charge.

From the question,

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Hence the magnitude of the electric field created by the

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