The large container shown in the cross section above is filled with a liquid of density 1.1 x 103 kg/m3 . a small hole of area 2
.5 x 10-6 m2 is opened in the side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into a beaker placed to the right of the container. at the same time, liquid is also added to the container at an appropriate rate so that h remains constant. the amount of liquid collected in the beaker in 2.0 minutes is 7.2 x 10-4 m3 . a. calculate the volume rate of flow of liquid from the hole in m3/s .
b. calculate the speed of the liquid as it exits from the hole.
c. calculate the height h of liquid needed above the hole to cause the speed you determined in part (b).
d. suppose that there is now less liquid in the container so that the height h is reduced to h/2. in relation to the beaker, where will the liquid hit the tabletop? left of the beaker in the
d. Imaginary point through which the resultant force of gravity acts on an object
Explanation:
Gravity acting on all particle points of the object, no matter how small they are. As a combination, the center of gravity would denote a single point which substitutes for all the gravity forces on the object.
To solve this problem we will apply the concept related to the electric potential energy, defined from the laws of Coulomb for which it is necessary to,
Here,
k = Coulomb's constant
= Charge of each object
r = Distance,
Our values are given as,
Finally replacing we have that,
Therefore the energy necessary to move the particle is 0.27J