An open cylindrical tank of acid rests at the edge of a table 1.4 m above the floor of the chemi

step 1 Okay, so in this problem we have a tank with exit. So let's draw here the tank. Let's put the ex

An open cylindrical tank of acid rests at the edge of a...

Key Concepts

Torricelli's Law

Torricelli's Law states that the exit speed of a fluid flowing from a small orifice under the influence of gravity is equal to the speed that a freely falling object would acquire over the same vertical distance. This relationship is given by the formula v = ?(2gh), where g is the acceleration due to gravity and h is the depth of the fluid above the opening. It is crucial for determining the initial horizontal velocity of the jet of fluid.

Projectile Motion

Projectile motion describes the two-dimensional motion of an object that is launched into the air and then moves under the influence of gravity alone, following a parabolic path. When the fluid exits the hole horizontally, it behaves like a projectile where the time to fall is determined solely by the vertical drop, while the horizontal distance traveled is a product of this time and the initial horizontal velocity.

Kinematic Equations

Kinematic equations provide the mathematical framework to analyze motion by relating displacement, velocity, acceleration, and time. In this context, they are used to calculate the time it takes for the fluid to fall from the edge of the table to the floor, and consequently, to determine the horizontal distance traveled by the jet of fluid using the previously determined exit velocity.

Transcript

00:01 Okay, so in this problem we have a tank with exit.

00:06 So let's draw here the tank.

00:09 Let's put the exit in the tank.

00:15 We know that this tank has a hole in one side from each the exit can flow.

00:22 Okay? so we know that the distance of the hole to the top of the exit is 75 centimeters.

00:37 Centimeters that is actually a bad 75 but it is 75 centimeters what else do you know we know that the tank lays above our table and the distance the height of this table is just let's put in here 1 .4 meters so this is 1 .4 meters we know that the trajectory of the motion of of the exit is going to be something like this.

01:13 And the problem just want to know what should be the distance from the table that the exit will land.

01:25 Ok, so first of all, let's remember that project our motion, we can separate the movement of the of the exit into coordinate system.

01:39 So in the x direction, we know that the motion of the exit is going to be determined by his initial speed in the x direction that multiplies the time of the fall.

01:53 In the y direction, we just have, let's see, we just have the acceleration, which is going to be the gravity acceleration, that multiplies t squared divided by 2.

02:11 So that is the motion for this case.

02:16 In the two coordinate system of this exit.

02:21 So since we want to calculate the x, the only problem here is that we have to calculate the time, and then we have to calculate the initial speed because we do not have neither the speed or the time.

02:38 So how we go into actually calculate this? well, let's keep this in here, and let's first of all calculate the speed we know that we can use since this is a tank with a column of liquid we can use the pressure the bernalise equation actually so we can say that that's called this point two and this point one so the bernalise equation is just let's say that let's say this point two and this point one so the bernalise equation is just let's write down here.

03:22 The bernouise equation is the pressure on point one plus the potential energy on point one, which is ho, g, y1, plus the connect energy, whole v1 square is going to be equal the pressure on point two plus hoh gy2 plus half of hoh.

03:55 V2 is square okay so first of all the pressure on point one needs to be equal the pressure on point two because we are in the same height so we can cross the pressure in here the speed on one is going to be zero because inside the tank the liquid the acid is not flowing so we can say this is 0.

04:30 Therefore, if we rearrange this equation, the speed of escape of the fluid is going to be v2 equals 2g.

04:44 They multiply y1 minus y2, all this in the square root.

04:55 So now we can finally calculate the speed.

04:58 Because this is speed 2 is going to be our initial speed here.

05:04 So let's see, this is just 2.

05:07 That multiplies 9 .8, which is the gravity acceleration.

05:14 They multiplies y1 minus y2, which is just 75 centimeters...