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nasty-shy [4]
3 years ago
7

A trapeze artist, with swing, weighs 800 N; he is momentarily heldto one side by his partner using a horizontal force so that th

eswing ropes make an angle of 30o with thevertical. In such a condition of static equilibrium, what isthe horizontal force being applied by the partner?

Physics
1 answer:
Tamiku [17]3 years ago
6 0

Answer:

461.88 N

Explanation:

F_{g} = Weight of the swing = 800 N

T = Tension force in the rope

F = Horizontal force being applied by the partner

Using equilibrium of force in vertical direction using the force diagram, we get

T Cos30 = F_{g}\\T Cos30 = 800\\T = \frac{800}{Cos30} \\\\T = 923.76 N

Using equilibrium of force in horizontal direction using the force diagram, we get

F = T Sin30\\F = (923.76) (0.5)\\F = 461.88 N

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Se lanza verticalmente hacia arriba un cuerpo que para por un punto A con una rapidez de 54 m/s y por otro punto B situado mas a
kvv77 [185]

Answer:

a) 3,06 seg

b) 119,39 m

Explanation:

<u>Lanzamiento Vertical </u>

Cuando un cuerpo se lanza verticalmente hacia arriba en el vacío, la única fuerza actuante es el peso. Si asumimos la dirección negativa hacia abajo, las fórmulas necesarias son

v_f=v_o+gt

y=y_o+\frac{gt^2}{2}

v_f^2=v_o^2+2gy

Siendo v_f la velocidad final, y la altura del objeto, g=-9.8\ m/seg^2 , v_o la velocidad inicial y t el tiempo

a)

Sabemos que el cuerpo pasa por un punto A a v_o=54\ m/s, y por otro punto B más arriba, a v_f=24\ m/s. El cuerpo está subiendo, pues pierde velocidad. Sabiendo las dos velocidades, podemos calcular el tiempo que toma en ir de A a B

\displaystyle t=\frac{v_f-v_o}{g}

\displaystyle t=\frac{24-54}{-9.8}

\displaystyle t=\frac{-30}{-9.8}=3,06\ seg

b)

Conociendo las velocidades de los extremos, se encuentra la distancia vertical que recorre durante ese intervalo

v_f^2=v_o^2+2gy

\displaystyle 24^2=54^2+2(-9.8)y

\displaystyle 24^2-54^2=-19.6y

Despejando y

\displaystyle y=\frac{576-2916}{-19.6}

y=119,39\ m

5 0
3 years ago
Before leaving an incident assignment, you should do all of the following EXCEPT FOR:
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<span>Self-dispatch to another incident</span>
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A car is initially driving at 30 m/s. It hits a large pothole, after which it is traveling in the same direction but at 25 m/s.
Bogdan [553]

Answer:

The time of Mars is 1.65 times larger on Mars than on Earth

Explanation:

The equation that describes the system is the final speed is equal to the speed minos the speed lost by the collision with the porhole

       Vf = Vo - V pothole

B) let's transform the weight of free groin system and N international system

      1 N = 0.2248 lb

      2.8 lbs (1N / 0.2248lbs) = 12.5 N

c) Kinematic equations are the same in all inertial systems, Mars and Earth, so we can use the height equation, with zero initial velocity

                   

        Y = Vo t - ½ g t²

        Y = - ½ g t²

        t = √ 2Y / g

     

Mars

         gm = 0.37g

         gm = 0.37 9.8

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         t = √( 2 1.9 / 3.626 )

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Earth

         t = √( 2 1.9 / 9.8)

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To make the comparison of time we are the relationship between the two

         tm / te = 1.02 / 0.62

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The time of Mars is 1.65 times larger on Mars than on Earth

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Answer:

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Explanation:

Evaporation takes place only at the surface of a liquid, whereas boiling may occur throughout the liquid. In boiling, the change of state takes place at any point in the liquid where bubbles form. The bubbles then rise and break at the surface of the liquid.

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