/*! 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 51 In a water pistol, a piston driv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a water pistol, a piston drives water through a larger tube of radius \(1.00 \mathrm{~cm}\) into a smaller tube of radius \(1.00 \mathrm{~mm}\) as in Figure \(\mathrm{P} 9.51\). (a) If the pistol is fired horizontally at a height of \(1.50 \mathrm{~m}\), use ballistics to determine the time it takes water to travel from the nozzle to the ground. (Neglect air resistance and assume atmospheric pressure is \(1.00 \mathrm{~atm}\).) (b) If the range of the stream is to be \(8.00 \mathrm{~m}\), with what speed must the stream leave the nozzle? (c) Given the areas of the nozzle and cylinder, use the equation of continuity to calculate the speed at which the plunger must be moved. (d) What is the pressure at the nozzle? (e) Use Bernoulli's equation to find the pressure needed in the larger cylinder. Can gravity terms be neglected? (f) Calculate the force that must be exerted on the trigger to achieve the desired range. (The force that must be exerted is due to pressure over and above atmospheric pressure.)

Short Answer

Expert verified
1) It takes approximately 0.55 seconds for water to hit the ground. 2) The water has to leave the nozzle at approximately 14.5 m/s to reach the target 8 meters away. 3) The plunger must be moved at a speed of approximately 14000 m/s. 4) The pressure at the nozzle is 1 atmosphere. 5) The pressure in the cylinder is approximately 1.1 atm, so gravity terms can be neglected. 6) The force that a person must exert on the trigger to achieve the desired range is approximately 1373 N over and above atmospheric pressure.

Step by step solution

01

Determine Time for Water to Hit the Ground

Use the kinematics equation \(h = \frac{1}{2} g t^2\), where \(h = 1.5 m\) is the height, \(g = 9.8 m/s^2\) is the acceleration due to gravity, and \(t\) is the time. Solving for \(t\) gives \(t = \sqrt{\frac{2h}{g}}\).
02

Determine Speed of Water Leaving the Nozzle

Use the kinematics equation \(d = v t\) to solve for \(v\), where \(d = 8 m\) is the distance and \(t\) is the time from step 1. This gives \(v = \frac{d}{t}\).
03

Determine Speed of the Piston

Use the equation of continuity \(A_1v_1 = A_2v_2\), where \(A_1\) is the area of the cylinder, \(v_1\) is the speed of the plunger, \(A_2\) is the area of the nozzle, and \(v_2\) is the speed of the water from step 2. Since the areas are \(\pi r^2\), the equation becomes \((\pi*1cm^2) v_1 = (\pi*1mm^2) v_2\), which simplifies to \(v_1 = \frac{1mm^2}{1cm^2} v_2\).
04

Determine Pressure at the Nozzle

The pressure at the nozzle is the atmospheric pressure, which is given as \(1 atm\). This is because the water is exposed to the atmosphere at this point.
05

Determine Pressure in the Cylinder

Use Bernoulli's principle \(p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2\), where \(p_1\) is the pressure in the cylinder, \(v_1\) is the speed of the plunger from step 3, \(\rho\) is the density of water, \(v_2\) is the speed of the water from step 2, and \(p_2\) is the pressure at the nozzle from step 4. Solving for \(p_1\) gives \(p_1 = p_2 + \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2\). The gravity terms can be neglected since the height change is negligible compared to the pressure difference.
06

Determine Force on the Pistol's Trigger

The force is the pressure difference times the piston's area, or \(F = (p_1 - p_2) A_1\), where \(A_1\) is the area of the cylinder, \(p_1\) is the pressure in the cylinder from step 5, and \(p_2\) is the atmospheric pressure.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Continuity Equation
In fluid mechanics, the continuity equation is a principle that shows how the flow rate of a fluid remains constant in a closed system. It's crucial for understanding how fluids behave as they move through different spaces. This principle is represented by the equation:
  • \( A_1 v_1 = A_2 v_2 \)
Here, \(A_1\) and \(A_2\) are the cross-sectional areas of the larger and smaller tube, respectively, and \(v_1\) and \(v_2\) are the fluid velocities at those points.
This equation implies that if a fluid enters a narrow space, its speed increases, just like water speeding up as it exits the narrow nozzle of a water pistol.
Understanding this helps in calculating how fast the piston must move to maintain a certain flow rate.
Bernoulli's Equation
Bernoulli's equation is an indispensable tool in fluid mechanics, derived from the principle of conservation of energy. It can be expressed as:
  • \[ p_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = p_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \]
-Where:
  • \(p_1\) and \(p_2\) are the pressures at two points along the streamline,
  • \(v_1\) and \(v_2\) are the fluid velocities,
  • \(\rho\) is the fluid's density,
  • \(g\) is the acceleration due to gravity, and
  • \(h_1\) and \(h_2\) are the heights of the fluid above some reference point.
In practical scenarios like the water pistol, this equation can be used to calculate the necessary pressure inside the cylinder to ensure the water exits the nozzle at a desired speed.
Significantly, in many cases, including this exercise, the gravity term can be neglected if the height difference is small compared to the pressure variation.
Kinematics
Kinematics here applies to the study of motion without considering the forces that cause this motion. To determine how long it takes water to hit the ground after leaving the nozzle, we use:
  • \( h = \frac{1}{2} g t^2 \)
This equation helps calculate the time, \(t\), it takes for the water to fall from the nozzle's height (1.5 m) to the ground. Solving for \(t\), we get:
  • \( t = \sqrt{\frac{2h}{g}} \)
Once the time is known, you can relate it to the range and determine the speed of the water leaving the nozzle.
This part of the problem requires an understanding of basic concepts of projectile motion, where the horizontal range is affected by initial velocity and time of flight.
Pressure Calculation
Pressure calculations in fluid mechanics often involve changes that occur when a fluid moves from one area to another. In our water pistol example, this means figuring out the pressures in the nozzle and the larger tube.
The pressure at the nozzle is the atmospheric pressure since this is the opening part of the system. However, inside the larger cylinder, the pressure differs, and it's calculated using Bernoulli's principle:
  • \( p_1 = p_2 + \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2 \)
This formula helps in understanding how pressure changes drive the flow through the system. Also, pressure difference helps us figure out the force needed on the trigger since this force directly correlates with the pressure needed to achieve the desired flow and range.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Whole blood has a surface tension of \(0.058 \mathrm{~N} / \mathrm{m}\) and a density of \(1050 \mathrm{~kg} / \mathrm{m}^{3}\). To what height can whole blood rise in a capillary blood vessel that has a radius of \(2.0 \times 10^{-6} \mathrm{~m}\) if the contact angle is zero?

On October 21, 2001, Ian Ashpole of the United Kingdom achieved a record altitude of \(3.35 \mathrm{~km}\) (11 \(000 \mathrm{ft}\) ) powered by 600 toy balloons filled with helium. Each filled balloon had a radius of about \(0.50 \mathrm{~m}\) and an estimated mass of \(0.30 \mathrm{~kg}\). (a) Estimate the total buoyant force on the 600 balloons. (b) Estimate the net upward force on all 600 balloons. (c) Ashpole parachuted to Earth after the balloons began to burst at the high altitude and the system lost buoyancy. Why did the balloons burst?

The aorta in humans has a diameter of about \(2.0 \mathrm{~cm}\), and at certain times the blood speed through it is about \(55 \mathrm{~cm} / \mathrm{s}\). Is the blood flow turbulent? The density of whole blood is \(1050 \mathrm{~kg} / \mathrm{m}^{3}\), and its coefficient of viscosity is \(2.7 \times 10^{-3} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\).

An object weighing \(300 \mathrm{~N}\) in air is immersed in water after being tied to a string connected to a balance. The scale now reads \(265 \mathrm{~N}\). Immersed in oil, the object appears to weigh \(275 \mathrm{~N}\). Find (a) the density of the object and (b) the density of the oil.

The total cross-sectional area of the load-bearing calcified portion of the two forearm bones (radius and ulna) is approximately \(2.4 \mathrm{~cm}^{2}\). During a car crash, the forearm is slammed against the dashboard. The arm comes to rest from an initial speed of \(80 \mathrm{~km} / \mathrm{h}\) in \(5.0 \mathrm{~ms}\). If the arm has an effective mass of \(3.0 \mathrm{~kg}\) and bone material can withstand a maximum compressional stress of \(16 \times 10^{7} \mathrm{~Pa}\), is the arm likely to withstand the crash?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.