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Scorpion4ik [409]
3 years ago
8

Help ASAP Just answer the first question for me please!

Physics
1 answer:
Usimov [2.4K]3 years ago
4 0

Answer:

Im pretty sure its b

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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
Why must 0 s always be clearly marked on a motion map?
Ainat [17]
The answer would be latter e
5 0
3 years ago
A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori
EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

x - x_o = u_xt

\mathtt{x = u_xt  \ \  \ since (x_o = 0)}  ---- (1)

the equation of the motion y is :

\mathtt{y - y_o =u_yt - 0.5 gt^2}

\mathtt{y = u_yt-4.9t^2     \ \ \  since (y_o =0)}

\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2        }

\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}

\mathtt{1 = 5.14t - 4.9t^2}

\mathtt{4.9t^2 - 5.14t +1 = 0}

By using the quadratic formula, we have;

\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}}     }

where;

a = 4.9,   b = -5.14     c = 1

= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}}     }

= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8}  \  \ OR \  \  \dfrac{ 5.14- \sqrt{6.8196}}{9.8}}    }

= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8}  \  \ OR \  \  \dfrac{ 5.14- 2.6114}{9.8}}    }

= \mathtt{ \dfrac{ 7.7514}{9.8}  \  \ OR \  \  \dfrac{ 2.5286}{9.8}}    }

= \mathbf{ 0.791 \  \ OR \  \  0.258}    }

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

\mathtt{x = u_x(0.258)}

\mathtt{x = ucos 40^0 (0.258)}

\mathtt{x = 8 \ cos 40^0 (0.258)}

\mathbf{x = 1.581  \ m}

Thus, the child is 1.581 m far from the fence

6 0
3 years ago
How do scientist determine how old the world is and the fossils inside of it?
valkas [14]

The layers of the fossil the oldest is usual the bottom layer,and the top layer is the newest layer

6 0
3 years ago
Read 4 more answers
A science teacher who was teaching a unit on minerals modeled mineral formation for the class. She heated granular sugar and a v
Olenka [21]

Answer:

Magnetite and quartz

Explanation:

5 0
3 years ago
Read 2 more answers
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