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3 years ago

Historia de baloncesto internacional resumen

Physics

1 answer:

xz_007 [3.2K]3 years ago

5 0

Answer:

El baloncesto tuvo sus inicios en el año de 1891, en los Estados Unidos de Norte América justamente en la ciudad de Springfield, el mismo surgió como alternativa al Hockey sobre hielo, el cual solo se realizaba durante la época del invierno. La YMCA fue la primera asociación en practicar baloncesto cuando su pionero y fundador era entrenador en dicha asociación.

Además de los jugadores, el equipo esta conformado por masajistas, medicos, Entrenador, director y asistente técnico, los cuales tienen funciones específicas para prestar asistencia y dirección a los jugadores.

Explanation:

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Answer:

Explanation:

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2 years ago

If we interpret the large, angular rocks to have originated from the outcrop at the top of the hill, we are using __________ rea

IrinaK [193]

The reasoning which is in use when large, angular rocks are interpreted to have originated from the outcrop at the top of the hill is; Fossil succession

<h3>Fossil succession of rocks</h3>

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On this note, any time period can be dated by its fossil content.

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2 years ago

Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light

SIZIF [17.4K]

Answer:

The correct answer is "4.443 sec".

Explanation:

Given:

Mass of child,

= 34 kg

Mass of swing,

= 18 kg

Length,

= 4.9 m

The time period of pendulum will be:

T = $2 \pi \sqrt{4g}$

= $2 \pi \sqrt{\frac{4.9}{9.8} }$

= $4.443 \ sec$

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3 years ago

Read 2 more answers

A car takes 10 s to go from vi=0 m/s to vf=25 m/s. How far does the car travel in<br> that time?

SSSSS [86.1K]

Answer:

2.5 k

Explanation:

8 0

3 years ago

The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of

Alenkasestr [34]

Answer:

a) $a=(-2,-6,0)m/s^2$ , with a magnitude of $6.3m/s^2$

b) $\frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2$ , with a magnitude of $0.76Kgm/s^2$

c) $F=(-0.24,-0.72,0)N$ , with a magnitude of $0.76N$

Explanation:

We have:

$v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg$

We can calculate each component of the acceleration using its definition $a=\frac{\Delta v}{\Delta t}$

$a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\$

The rate of change of momentum of the ball is $\frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma$

So for each coordinate:

$\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2$

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

$a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N$

3 0

3 years ago