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RSB [31]
3 years ago
15

Course hero N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches

the bottom side of the board at a point 85 cm from the end under the rock. The board is 4.5 m long, has a mass of about 22 kg, and projects essentially horizontally out over a river. Is it safe for an adult with a mass of 62 kg to stand at the unsupported end of the board
Physics
1 answer:
aleksandrvk [35]3 years ago
8 0

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass m=380kg

Height of supporting Rock X=85cm

Length of BoardL_r=4.5m

Mass of board M_b=22kg

Mass of adult M_a=62  

Generally the moment of balance about wedge part about  is mathematically given by

N -Q + R = Mg + mg

0.85*N - Mg*2.25 - mg*(2.25 + x) = 0

0.85*N  = + Mg*2.25 + mg*(2.25 + x)

where

N+R=4547

therefore

N = 570.70588 + 1608.3529 + 714.823 x

if N=0 at fallen person

x=3.04m

Therefore it is save to carry a 62kg adult

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a) Las balas de cañon disparadas desde el puerto deben tener una ángulo de 27.39° para que puedan impactar al barco, con una velocidad inicial de 82 m/s.

b) El tiempo de vuelo de las balas de cañon para alcanzar al barco que está a 560 m de distancia es de 7.69 s.

c) Sabiendo que el ángulo calculado en el inciso a) es para una distancia de 560 m, el barco debe estar a una distancia mayor para que las balas no lo alcancen.

a)

Podemos usar las ecuaciones de tiro parabólico para encontrar el ángulo que permita derribar al barco invasor.

x=\frac{v_{i}^{2}sin(2\alpha)}{g} (1)

Donde:

  • v(i) es la velocidad inicial del cañon (82 m/s)
  • α es el ángulo de tiro
  • g es la gravedad (9.81 m/s²)
  • x es el desplazamiento total (560 m)

Lo que debemos hacer es depejar α de la ecuación 1

sin(2\alpha)=\frac{xg}{v_{i}^{2}}

sin(2\alpha)=\frac{560*9.81}{82^{2}}

sin(2\alpha)=0.82

\alpha=\frac{sin^{-1}(0.82)}{2}

\alpha=27.39^{\circ}

Por lo tanto, el ángulo para que el cañón impacte en el barco es de 27.39 °.

b)

Sabemos que la componente de la velocidad en el eje x es constante, así que podemo usar la siguiente ecuación.

v_{x}=\frac{x}{t}

La componente x de la velocidad es V(x) = V(i)cos(α) y sabiendo la distancia total de 560 m, el tiempo será:

v_{i}cos(\alpha)=\frac{x}{t}

t=\frac{x}{v_{i}cos(\alpha)}

t=\frac{560}{82cos(27.39)}

t=7.69\: s

El tiempo total de vuelo de las balas de cañon es de 7.69 s.

c)

Sabemos que el ángulo calculado en el inciso a) es de 27.39 °, y ese valor fue calcualdo para una distancia de 560 m, por lo tanto el barco debe estar a una distancia mayor que esa para que las balas no lo alcancen.

Puedes encontrar más información sobre tiro parabólico aquí:

https://brainly.lat/tarea/3605927

Espero te haya sido de ayuda!

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Answer:

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Explanation:

Total pressure is constant and PT = P = 1/2×ρ×v^2  

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A2 = 4×A1  

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