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2 years ago

Historia de baloncesto internacional resumen

Physics

1 answer:

xz_007 [3.2K]2 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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Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$

so, 4 times the original acceleration

- We can decrease the radius to 1/4 of its original value:

$r'=\frac{1}{4}r$

So the new acceleration is

$a'=\frac{v^2}{(r/4)}=4(\frac{v^2}{r})=4a$

so, the acceleration has increased by a factor 4 again.

Learn more about centripetal acceleration:

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3 years ago

What is the speed of terminal velocity?

ivann1987 [24]

The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.

A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.

It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:

-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .

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3 years ago

A 160.-kilogram space vehicle is traveling along a straight line at a constant speed of 800. Meters per second. The magnitude of

natulia [17]

Answer:

Zero

Explanation:

As force acting on the body is equal to the product of mass and acceleration.

Acceleration is equal to rate of change in velocity.

Here velocity is constant so acceleration is zero.

It means the net force acting on the vehicle is zero.

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3 years ago

Read 2 more answers

As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change

Umnica [9.8K]

Refer to the diagram shown below.

m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
x is zero;
v is maximum;
a is zero.

At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.

In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.

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3 years ago

1.When you give one set of washers a downward push, does it move as easily as the other set? Does it stop before it reaches the

Vlad1618 [11]

Answer:No, it doesn't move easily downward because it will try to resist the movement ,due to a resistance force of inertia that it possess at rest.

Explanation:when an object has higher or larger mass it tends to resist any motion given to it unlike the one with lower mass.

The larger the mass the more resistance force an object has.

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3 years ago