The total gauge pressure at the bottom of the cylinder would
simply be the sum of the pressure exerted by water and pressure exerted by the
oil.
The formula for calculating pressure in a column is:
P = ρ g h
Where,
P = gauge pressure
ρ = density of the liquid
g = gravitational acceleration
h = height of liquid
Adding the two pressures will give the total:
P total = (ρ g h)_water + (ρ g h)_oil
P total = (1000 kg / m^3) (9.8 m / s^2) (0.30 m) + (900 kg /
m^3) (9.8 m / s^2) (0.4 - 0.30 m)
P total = 2940 Pa + 882 Pa
P total = 3,822 Pa
Answer:
The total gauge
pressure at the bottom is 3,822 Pa.
Answer:
The compression is
.
Explanation:
A Hooke's law spring compressed has a potential energy

where k is the spring constant and
the distance to the equilibrium position.
A mass m moving at speed v has a kinetic energy
.
So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity
. Knowing that the energy is constant.

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

But, in the left side we can use the previous equation to obtain:





And this is the compression we are looking for
Answer:
Ben's average speed was twice Debby's average speed.
Explanation:
Ben covered a total distance of 16 miles (10+4+2) and Debby covered 8 miles (3+2+2+1) which is half of what Ben covered. As they both reached the place in the same amount of time it tells us Ben was faster.
Answer:
My scenario would be A Car vs. a guard rail on a road. You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)
And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.
Explanation:
This is caused by so much force reacting from one object to another but also depends on molecular density.
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.