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3 years ago

At what point in its path does a baseball that is hit to the outfield have its minimum vertical speed?

Physics

1 answer:

ioda3 years ago

5 0

Answer:

At the point when it reaches its maximum height (answer C)

Explanation:

As soon as the ball leaves the bat is has not only horizontal component of the velocity, but also vertical, which allows it travel upwards in its trajectory.

When it reaches its maximum elevation, and the trajectory starts curving down, is when the velocity of the baseball is just horizontal (the vertical component has been reduced to zero due to the constant action of the acceleration of gravity which was slowing down it ascending pattern.

After that point, the ball starts gaining now vertical velocity being accelerated towards the center of the Earth due to gravity. At the maximum height of its path is when the vertical component of the velocity switches from positive (pointing up) to negative (pointing down) going through zero magnitude.

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A 0.111 kg hockey puck moving at 55 m/s is caught by a 80 kg goalie at rest. With what speed does the goalie slide on the (frict

Andrews [41]

Answer: 0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

6 0

3 years ago

Como anticipan la caída del proyectil, con movimiento parabólico para que el misil no haga daño

Elena L [17]

Answer:

Conociendo la velocidad inicial del proyectil y el angulo de lanzamiento con respecto ala horizontal.

Explanation:

Para poder anticipar la caída del proyectil es importante conocer la velocidad inicial del proyectil y el angulo de disparo del proyectil con respecto a la horizontal.

A continuación se presenta un diagrama o esquema donde se pueden ver estas variables y se explicaran a la brevedad:

Para poder encontrar el rango que es la máxima distancia horizontal recorrida por el proyectil debemos utilizar la siguiente ecuación:

$x=(v_{o})_{x} *t\\where:\\(v_{o})_{x} = velocidad inicial x-component [m/s]\\t= time [s]$

Para poder encontrar el tiempo debemos utilizar la siguiente ecuación:

$y=(v_{y} )_{o}*t-0.5*g*t^{2} \\donde:\\(v_{y} )_{o}= velocidad inicial componente y [m/s]\\g = gravity = 9.81 [m/s^2]\\t = time [s]$

En la anterior ecuación, igualamos y = 0, ya que cuando el proyectil cae al suelo la distancia vertical es cero. De esta manera podemos encontrar el tiempo t, ya que conocemos la velocidad inicial del proyectil en la componente y.

Seguidamente reemplazamos t en la primera ecuacion y encontramos la distancia x o el rango.

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4 years ago

A proton of mass 1.67×10-27 kg is being accelerated along a straight line at 7.01×1015 m/s2 in an accelerator. If the proton has

Masja [62]

Answer:

$v_f=7*10^7\frac{m}{s}$

Explanation:

The proton is under a linear motion with constant acceleration. So, we use the kinemtic equations to calculate its final speed. We know its acceleration, its initial speed and its traveled distance. Thus, we use the following equation:

$v_f^2=v_0^2+2ax\\v_f=\sqrt{v_0^2+2ax}\\v_f=\sqrt{(6.63*10^7\frac{m}{s})^2+2(7.01*10^{15}\frac{m}{s^2})(3.63*10^{-2}m)}\\v_f=7*10^7\frac{m}{s}$

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4 years ago

________________ is the tendency of an object to resist a change in its motion.

Sliva [168]

Inertia is the tendency of an object to resist a change in its motion.

7 0

3 years ago

A sled is initially given a shove up a frictionless 37.0 ∘ incline. It reaches a maximum vertical height 1.20 m higher than wher

labwork [276]

The initial speed of the sled at the given height is 4.85 m/s.

<h3>Initial speed of the sled</h3>

Apply the principle of conservation of energy;

K.E = P.E

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

h is the vertical height reached
g is acceleration due to gravity

v = √(2 x 9.8 x 1.2)

v = 4.85 m/s

Thus, the initial speed of the sled at the given height is 4.85 m/s.

Learn more about initial speed here: brainly.com/question/24493758

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5 0

2 years ago