1. Se dispara una bala de 10grbcon una velocidad de 500m/s contra un muro de 10cm de espesor. Si la resistencia del muro al avance de la bala es de 3000 N, calcula la velosidad calcula la velocidad de la bala después de atravesar el muro. 2. Un automóvil de 1000kg de masa aumenta su velosidad de 0 a 100 km/h en un tiempo mínimo de 8s calcula su potencia en watios y en caballos de vapor. Dato : 1cv = 735w. 3. Desde una altura de 10m deja caer un cuerpo de 5kg. Calcula su velocidad al llegar al suelo

Answers

Answer 1

Answer:

1) La velocidad de la bala después de atravesar el muro es de aproximadamente 435,890 metros por segundo.

2) La potencia del automóvil es 96438,272 watts o 131,208 caballos de vapor.

3) La velocidad del objeto al llegar al suelo es aproximadamente 14,005 metros por segundo.

Step-by-step explanation:

1) La velocidad final de la bala puede determinarse mediante el Teorema del Trabajo y la Energía, a partir del cual se tiene la siguiente fórmula:

[tex]\frac{1}{2}\cdot m\cdot v_{o}^{2} -F\cdot \Delta s = \frac{1}{2}\cdot m \cdot v_{f}^{2}[/tex] (1)

Where:

[tex]m[/tex] - Masa de la bala, en kilogramos.

[tex]v_{o}, v_{f}[/tex] - Velocidades inicial y final de la bala, en metros por segundo.

[tex]F[/tex] - Resistencia del muro al avance de la bala, en newtons.

[tex]\Delta s[/tex] - Espesor del muro, en metros.

Si sabemos que [tex]m = 0,01\,kg[/tex], [tex]v_{o} = 500\,\frac{m}{s}[/tex], [tex]F = 3000\,N[/tex] and [tex]\Delta s = 0,1\,m[/tex], entonces la velocidad final de la bala es:

[tex]v_{f}^{2}=v_{o}^{2} -\frac{2\cdot F\cdot \Delta s}{m}[/tex]

[tex]v_{f} = \sqrt{v_{o}^{2}-\frac{2\cdot F\cdot \Delta s}{m} }[/tex]

[tex]v_{f} \approx 435,890\,\frac{m}{s}[/tex]

La velocidad de la bala después de atravesar el muro es de aproximadamente 435,890 metros por segundo.

2) Asumamos que el automóvil acelera a tasa constante, significando que la fuerza neta será constante. Para un sistema cuya fuerza neta sea constante, la potencia experimentada queda descrita por la siguiente ecuación:

[tex]P = m\cdot a(t)\cdot v(t)[/tex] (2)

[tex]a(t) = a[/tex] (3)

[tex]v(t) = v_{o} + a\cdot t[/tex] (4)

Donde:

[tex]P[/tex] - Potencia, en watts.

[tex]m[/tex] - Masa del automóvil, en kilogramos.

[tex]a(t)[/tex] - Aceleración, en metros por segundo al cuadrado.

[tex]v(t)[/tex] - Velocidad, en metros por segundo.

[tex]v_{o}[/tex] - Velocidad inicial del automóvil, en metros por segundo.

Si sabemos que [tex]m = 1000\,kg[/tex], [tex]a = 3,472\,\frac{m}{s}[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex] y [tex]t = 8\,s[/tex] entonces la potencia experimentada por el automóvil es:

[tex]P = 96438,272\,W[/tex] ([tex]131,208\,C.V.[/tex])

La potencia del automóvil es 96438,272 watts o 131,208 caballos de vapor.

3) El cuerpo experimenta un Movimiento de Caída Libre, el cual es un Movimiento Uniformemente Acelerado debido a la gravedad terrestre. La velocidad del cuerpo al llegar al suelo se determina mediante la siguiente fórmula cinemática:

[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot g\cdot h}[/tex] (5)

Donde:

[tex]v_{o}[/tex] - Velocidad inicial del cuerpo, en metros por segundo.

[tex]v_{f}[/tex] - Velocidad final del cuerpo, en metros por segundo.

[tex]g[/tex] - Aceleración gravitacional, en metros por segundo al cuadrado.

[tex]h[/tex] - Altura recorrida por el cuerpo, en metros.

Si sabemos que [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = 9,807\,\frac{m}{s^{2}}[/tex] y [tex]h = 10\,m[/tex], entonces la velocidad al llegar al suelo es:

[tex]v_{f} \approx 14,005\,\frac{m}{s}[/tex]

La velocidad del objeto al llegar al suelo es aproximadamente 14,005 metros por segundo.


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Answers

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Answers

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