Answer:
in 90 seconds It'll travel 450m At constant speed of 5m/s
Answer:
Neural Engineering and Tissue Engineering
Explanation:
• BME is Biomedical Engineering. It has various topics such as Anantomy, Physiology, Neural Engineering and Tissue engineering, Medical mechanisation and others.

Answer:Momentum
Explanation:
Product of an object mass and velocity is called Momentum. Basically momentum is the measure of mass in motion . More is the mass more will be the momentum and similar applies for velocity.
For example a bus moving with low velocity can have a same momentum as that of a bullet moving with a very high velocity
Momentum is denoted by Symbol P

Answer:
<em>The gravitational force between Royce and Earth would be doubled at 16 years.</em>
Explanation:
<em>"Newton's law of universal gravitation states that gravitation force between two masses is proportional to the magnitude of their masses and inverse-squared of their distance".</em>
Royce Scenario
At the age of 10 Royce's mass = 30kg
At the age of 16 Royce's mass = 60kg
From Newton's law of universal gravitation, an Increase in the mass of a body would amount to a corresponding increase in the gravitational force.
In the case of Royce, the mass double between the age of 10 and 16, so there would be an increase of the gravitation force by double.
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]](https://tex.z-dn.net/?f=x%3D%28v_%7Bo%7D%29_%7Bx%7D%20%2At%5C%5Cwhere%3A%5C%5C%28v_%7Bo%7D%29_%7Bx%7D%20%3D%20velocidad%20inicial%20%20x-component%20%5Bm%2Fs%5D%5C%5Ct%3D%20time%20%5Bs%5D)
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]](https://tex.z-dn.net/?f=y%3D%28v_%7By%7D%20%29_%7Bo%7D%2At-0.5%2Ag%2At%5E%7B2%7D%20%20%5C%5Cdonde%3A%5C%5C%28v_%7By%7D%20%29_%7Bo%7D%3D%20velocidad%20inicial%20componente%20y%20%5Bm%2Fs%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%20%5Bm%2Fs%5E2%5D%5C%5Ct%20%3D%20time%20%5Bs%5D)
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.