/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 A gallon of milk in a full plast... [FREE SOLUTION] | 91Ó°ÊÓ

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A gallon of milk in a full plastic jug is sitting on the edge of your kitchen table. Estimate the vertical distance between the top surface of the milk and the bottom of the jug. Also estimate the distance from the tabletop to the floor. You punch a small hole in the side of the jug just above the bottom of the jug, and milk flows out the hole. When the milk first starts to flow out the hole, what horizontal distance does it travel before reaching the floor? Assume the milk is in free fall after it has passed through the hole, and neglect the viscosity of the milk.

Short Answer

Expert verified
The horizontal distance \(d\) the milk will travel before reaching the floor is given by \(\sqrt{4d_td_f}\). If \(d_t=0.1m\) and \(d_f=1m\), the distance is approximately 0.632m or 63.2cm.

Step by step solution

01

Estimating the vertical distance

Firstly, estimate the vertical distance from the top surface of the milk to the bottom of the jug and from the tabletop to the floor. Let's say it is \(d_t\) with a typical value of 0.1m and \(d_f\) with a typical value 1m respectively.
02

Application of Torricelli's theorem

Next, apply Torricelli's theorem which states that the speed \(v\) of efflux of a fluid under the force of gravity from a small hole located \(h\) under the surface of the fluid is given by \(v=\sqrt{2gh}\). Here, \(g\) is the acceleration due to gravity and \(h\) is the depth of fluid above the hole. Hence, \(v=\sqrt{2gd_t}\).
03

Applying kinematics

Once the milk is out of the hole, it's in free fall. From kinematics, the time it takes to reach the floor \(t\) is given by \(\sqrt{2d_f / g}\).
04

Calculating horizontal distance traveled by milk

Finally, the horizontal velocity of the milk is the same as its initial velocity, which is the exit velocity from the hole. So, the horizontal distance \(d\) the milk travels before reaching the floor is given by \(d=vt = \sqrt{2gd_t} \* \sqrt{2d_f / g}= \sqrt{4d_td_f}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics deals with the motion of objects without considering the forces that cause the motion. It's essential for understanding how the milk travels after leaving the jug in our problem. The milk stream follows a path as it moves horizontally and vertically to the floor after escaping the jug. This movement falls under kinematic analysis because it involves calculating the distance, speed, and time of travel for the milk.

In this scenario, once the milk exits the jug, it is essentially in projectile motion even though it's a downward motion. Using kinematic equations, we can predict how far the milk travels horizontally before hitting the ground. Knowing the time from when the milk leaves the jug until it hits the floor is crucial. By using these simple kinematic equations, we simplify how we understand motions in one or two dimensions.
Fluid Dynamics
Fluid dynamics is the study of fluids in motion. It involves understanding how liquids and gases move and the forces they encounter. In this exercise, fluid dynamics helps us understand how the milk flows out from a small hole in the jug. One fundamental principle used is Torricelli's Theorem:
  • Torricelli's theorem helps determine the speed at which fluid exits a hole due to gravity.
  • It states: \( v = \sqrt{2gh} \), where \( v \) is the velocity of the efflux, \( g \) is the gravity acceleration, and \( h \) is the height from the fluid surface to the hole.
When you punch a hole at the bottom of the jug, the milk starts to flow out at a speed determined by the height of the milk above the hole. Fluid dynamics enables us to predict and calculate how fast the milk exits leading to understanding the flow characteristics of milk.
Free Fall
Free fall is a type of motion that theoretically happens when an object is acted upon only by gravity. When the milk leaves the jug, it enters free fall because all the horizontal supporting forces vanish, and only gravity pulls it down.

Studying free fall helps us calculate how quickly objects accelerate due to gravity, which is approximately \(9.8 \, \text{m/s}^2\) on Earth. Because the milk is in free fall after leaving the jug, it moves vertically downward under gravity's influence alone, while maintaining its horizontal velocity gained from exiting the jug.

Understanding free fall is key here because it allows us to accurately estimate how long it takes for the milk to hit the ground once it starts falling. This period determines the time the milk has to travel horizontally, directly affecting how far it goes before reaching the floor.

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Most popular questions from this chapter

Viscous blood is flowing through an artery partially clogged by cholesterol. A surgeon wants to remove enough of the cholesterol to double the flow rate of blood through this artery. If the original diameter of the artery is \(D,\) what should be the new diameter (in terms of \(D\) ) to accomplish this for the same pressure gradient?

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On the afternoon of January \(15,1919,\) an unusually warm day in Boston, a 17.7 -m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m-deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of \(1600 \mathrm{~kg} / \mathrm{m}^{3}\). If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (Hint: Consider the outward force on a circular ring of the tank wall of width \(d y\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

A cylindrical disk of wood weighing \(45.0 \mathrm{~N}\) and having a diameter of \(30.0 \mathrm{~cm}\) floats on a cylinder of oil of density \(0.850 \mathrm{~g} / \mathrm{cm}^{3}\) (Fig. E12.19). The cylinder of oil is \(75.0 \mathrm{~cm}\) deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column? (b) Suppose now that someone puts a weight of \(83.0 \mathrm{~N}\) on top of the wood, but no oil seeps around the edge of the wood. What is the change in pressure at (i) the bottom of the oil and (ii) halfway down in the oil?

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