/*! 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 33 (III) A scale is adjusted so tha... [FREE SOLUTION] | 91Ó°ÊÓ

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(III) A scale is adjusted so that when a large, shallow pan is placed on it, it reads zero. A water faucet at height \(h=2.5 \mathrm{~m}\) above is turned on and water falls into the pan at a rate \(R=0.14 \mathrm{~kg} / \mathrm{s} .\) Determine \((a)\) a formula for the scale reading as a function of time \(t\) and \((b)\) the reading for \(t=9.0 \mathrm{~s} .(c)\) Repeat \((a)\) and \((b),\) but replace the shallow pan with a tall, narrow cylindrical container of area \(A=20 \mathrm{~cm}^{2}\) (the level rises in this case).

Short Answer

Expert verified
(a) \( W(t) = 0.14tg \), (b) \( W(9) = 12.37 \, N \), (c) \( W(t) = R\sqrt{2gh} + 0.14tg \), (d) \( W(9) = 13.35 \, N \).

Step by step solution

01

Understanding the Problem Setup

We have a scale that initially reads zero when a pan is placed on it. Water falls into this pan from a height of \( h=2.5 \, \text{m} \) at a rate of \( R=0.14 \, \text{kg/s} \). We need to find how the scale reading changes over time as water collects in the pan.
02

Calculating Scale Reading for Part (a)

For a pan, the water in it directly affects the scale reading. Since water is added at a constant rate, the mass of water after time \( t \) is \( M = R \, t = 0.14 \, \text{kg/s} \times t \). The scale reads the weight, which is mass times gravitational acceleration \( g \). Therefore, the scale reading \( W \) as a function of time is \( W(t) = M \, g = 0.14 \, t \, g \).
03

Numerical Calculation for Part (b)

Given \( t = 9 \, \text{s} \), we find the scale reading using \( g = 9.81 \, \text{m/s}^2 \), so \( M = 0.14 \times 9 \). Thus, \( W(9) = 0.14 \times 9 \times 9.81 \approx 12.37 \, \text{N} \).
04

Adjusting for a Tall, Narrow Container in Part (c)

For the cylindrical container, water accumulates from the bottom. We need to consider the hydrostatic pressure affecting the scale. The effective weight on the scale now includes both the dynamic impact of incoming water and the accumulated static water weight. Calculating this requires the dynamic fluid impact force formula \( F = R v \) where \( v = \sqrt{2gh} \) and gravity's effect from accumulated water mass \( M g \). Thus: \( W(t) = R \sqrt{2gh} + 0.14 \, t \, g \).
05

Calculation for Part (d)

At \( t = 9 \, \text{s} \) for the cylinder, using \( R = 0.14 \, \text{kg/s} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 2.5 \, \text{m} \):\[ v = \sqrt{2 \times 9.81 \times 2.5} \approx 7 \, \text{m/s} \]Thus,\[ F = 0.14 \, \times 7 + 0.14 \times 9 \times 9.81 \approx 0.98 + 12.37 = 13.35 \, \text{N} \].

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Key Concepts

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

Hydrostatics
Hydrostatics is the branch of physics concerned with the study of fluids at rest. When dealing with water or any fluid in a stationary state, it is important to understand the concept of pressure and how it relates to depth and area.
In this particular exercise, we are asked to assess how the water’s pressure affects the reading on a scale. Pressure in a fluid is due to its weight, and it increases with depth. The basic equation for hydrostatic pressure is given by:
  • Pressure,
    \[ P = ho g h \]
    where \( P \) is the pressure, \( \rho \) is fluid density, \( g \) is acceleration due to gravity, and \( h \) is the height of the fluid column.
In our case, when using a cylindrical container, the level of accumulated water affects the pressure at the bottom, and hence the measurement of weight on the scale. Understanding this helps in calculating the total force exerted on the container bottom due to the water column.
Weight Calculation
When water is falling into the pan, the key point is understanding how weight changes with time. Weight is the force exerted by gravity on a mass. The formula for weight when concerning water falling into a container is:
  • Weight,
    \[ W(t) = M \, g \]
    where \( M \) is the mass of water collected over time \( t \), and \( g \) is gravitational acceleration.
This problem involves water being added at a constant rate, so the mass \( M \) increases linearly with time, resulting in a linear increase in weight on the scale.
For instance, after 9 seconds, the mass of water collected in the pan is calculated by multiplying the rate of water input \( R \) by time, and then by gravity to obtain the weight. This calculated weight corresponds to the scale reading.
Dynamics of Fluids
The dynamics of fluids involve understanding how moving fluids exert forces on objects. This part of physics deals with the behavior of liquids in motion, and in this case, it includes both static and dynamic forces.
When focusing on the exercise, two key aspects must be considered: the static weight of accumulated water and the dynamic impact of incoming water.
For the dynamic impact, one must understand how the velocity of falling water influences the force it exerts on the container. The impact force can be calculated using:
  • Dynamic Force,
    \[ F = R \, v \]
    where \( F \) is the force, \( R \) is the rate of water flow, and \( v \) is the velocity of water just before impact. The velocity \( v \) is found using the formula \( v = \sqrt{2gh} \) where \( h \) is the height from which the water falls.
This dynamic impact must be added to the static weight to find the total force exerted on the scale. For a comprehensive understanding, combining these aspects allows for precise calculation of how fluid dynamics influence the scale's reading at any given time.

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Most popular questions from this chapter

(II) An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.5 times the mass of the other. If \(7500 \mathrm{~J}\) is released in the explosion, how much kinetic energy does each piece acquire?

(II) The jet engine of an airplane takes in \(120 \mathrm{~kg}\) of air per second, which is burned with \(4.2 \mathrm{~kg}\) of fuel per second. The burned gases leave the plane at a speed of \(550 \mathrm{~m} / \mathrm{s}\) (relative to the plane). If the plane is traveling \(270 \mathrm{~m} / \mathrm{s}(600 \mathrm{mi} / \mathrm{h}),\) determine: \((a)\) the thrust due to ejected fuel; \((b)\) the thrust due to accelerated air passing through the engine; and ( \(c\) ) the power (hp) delivered.

Two bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear (Fig. \(9-56\) ). Car A has a mass of \(450 \mathrm{~kg}\) and car \(\mathrm{B} 490 \mathrm{~kg}\), owing to differences in passenger mass. If car A approaches at \(4.50 \mathrm{~m} / \mathrm{s}\) and car \(\mathrm{B}\) is moving at \(3.70 \mathrm{~m} / \mathrm{s},\) calculate \((a)\) their velocities after the collision, and \((b)\) the change in momentum of each.

(III) For an elastic collision between a projectile particle of mass \(m_{\mathrm{A}}\) and a target particle (at rest) of mass \(m_{\mathrm{B}}\), show that the scattering angle, \(\theta_{\mathrm{A}}^{\prime},\) of the projectile \((a)\) can take any value, 0 to \(180^{\circ},\) for \(m_{\mathrm{A}}m_{\mathrm{B}}\).

(III) A 3.0-kg block slides along a frictionless tabletop at \(8.0 \mathrm{~m} / \mathrm{s}\) toward a second block (at rest) of mass \(4.5 \mathrm{~kg} . \mathrm{A}\) coil spring, which obeys Hooke's law and has spring constant \(k=850 \mathrm{~N} / \mathrm{m},\) is attached to the second block in such a way that it will be compressed when struck by the moving block, Fig. \(9-40 .\) ( \(a\) ) What will be the maximum compression of the spring? (b) What will be the final velocities of the blocks after the collision? \((c)\) Is the collision elastic? Ignore the mass of the spring.

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