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Agata [3.3K]
4 years ago
11

The atomic number of an atom is the number of ____.

Physics
2 answers:
Anestetic [448]4 years ago
6 0

It is for sure protons i just took the test.

NISA [10]4 years ago
3 0
D protons

k12 student here
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Se lanza desde el suelo verticalmente hacia arriba un objeto con una velocidad inicial de 20 m/s. Calcula: a)Posición y velocida
djverab [1.8K]

Answer:

Ver las respuestas a continuacion.

Explanation:

Para solucionar este tipo de problemas debemos de utilizar la siguiente ecuacion de la cinematica.

a)

v_{f}=v_{o}-g*t

Donde:

Vf = velocidad final [m/s]

Vo = velocidad inicial = 20 [m/s]

g = aceleración gravitacional = 9.81 [m/s²]

t = tiempo = 1 [s]

El signo negativo de la aceleración gravitacional significa que el objeto se mueve en dirección contraria a la aceleración gravitacional.

v_{f}=20-9.81*(1)\\v_{f}=10.19[m/s]

b)

La altura máxima se logra cuando la velocidad final es cero, es decir no existe movimiento alguno para que siga subiendo el objeto. Debemos utilizar la siguiente ecuación de la cinemática.

v_{f}^{2} =v_{o}^{2}-2*g*y

Donde:

y = elevación [m]

0 = 20^{2} -2*9.81*y\\19.62*y=400\\y=20.38 [m]

c)

Podemos calcular el tiempo máximo que toma el objeto en subir.

v_{f}=v_{o}-g*t

0 = 20 -9.81*t\\9.81*t=20\\t=2.04[s]

Ahora del numeral b) sabemos la distancia maxima, de esta manera y utilizando la siguiente ecuacion podemos calcular el tiempo de caida, Debemos tener en cuenta que ahora la velocidad final es igual a la velocidad inicial = 0.

y=y_{o}+v_{o}*t+\frac{1}{2}*g*t^{2}

20.38 = 0*t + 0.5*9.81*t^{2} \\20.38 = 4.905*t^{2}\\t=\sqrt{\frac{20.38}{4.905} } \\t=2.04[s]

Entonces el tiempo total es igual a:

t = 2.04 + 2.04

t = 4.076 [s]

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