/*! 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 73 The Force of a Storm During a se... [FREE SOLUTION] | 91Ó°ÊÓ

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The Force of a Storm During a severe storm in Palm Beach, \(\mathrm{FL}\), on January 2,1999,31 inches of rain fell in a period of nine hours. Assuming that the raindrops hit the ground with a speed of \(10 \mathrm{m} / \mathrm{s},\) estimate the average upward force exerted by one square meter of ground to stop the falling raindrops during the storm. (Note: One cubic meter of water has a mass of \(1000 \mathrm{kg}\)

Short Answer

Expert verified
The average upward force exerted by the ground is approximately 0.243 N per square meter.

Step by step solution

01

Convert Rainfall to Meters

First, we need to convert the rainfall from inches to meters. Since 1 inch is 0.0254 meters, we multiply the rainfall amount by this conversion factor:\[31 \text{ inches} \times 0.0254 \text{ meters/inch} = 0.7874 \text{ meters}\]This means 0.7874 meters of rain fell during the storm.
02

Calculate the Volume of Water per Square Meter

Next, determine the volume of water that fell on one square meter of ground. Since the rainfall was 0.7874 meters high, the volume per square meter is:\[V = 1 \text{ m}^2 \times 0.7874 \text{ m} = 0.7874 \text{ m}^3\]Thus, the volume on one square meter is 0.7874 cubic meters.
03

Determine the Mass of Water

The mass of a volume of water can be calculated since 1 cubic meter of water has a mass of 1000 kg. For 0.7874 cubic meters, the mass is:\[m = 0.7874 \text{ m}^3 \times 1000 \text{ kg/m}^3 = 787.4 \text{ kg}\]So, the mass of rainwater on one square meter is 787.4 kg.
04

Calculate the Change in Momentum

Calculate the change in momentum (impulse) needed to stop the rain. Assuming the rain hits with a velocity of 10 m/s and stops at the ground, the change in momentum is:\[p = m \times v = 787.4 \text{ kg} \times 10 \text{ m/s} = 7874 \text{ kg}\cdot\text{m/s}\]This is the impulse experienced by one square meter of ground.
05

Calculate the Time of the Storm

Convert the storm duration from hours to seconds. Since 9 hours is equal to 9 x 60 x 60 seconds:\[9 \times 3600 = 32400 \text{ seconds}\]The storm lasted for 32400 seconds.
06

Calculate the Average Force Exerted

Using the formula for force: \( F = \frac{\text{Impulse}}{\text{time}} \), we find the average force:\[F = \frac{7874 \text{ kg}\cdot\text{m/s}}{32400 \text{ s}} = 0.243 \text{ N}\]The average upward force per square meter is approximately 0.243 Newtons.

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Key Concepts

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

Impulse and momentum
Impulse and momentum are key concepts in physics, especially when analyzing forces and collisions. When an object experiences a change in velocity, its momentum changes, and the impulse is the measure of that change. The impulse given to an object can be calculated using the equation:\[ \text{Impulse} = \Delta p = m \times \Delta v \]where \( \Delta p \) is change in momentum, \( m \) is the mass and \( \Delta v \) is the change in velocity.
In the context of our storm example, the raindrops had an initial velocity as they fell towards the ground. When they hit the ground, they came to a stop, which means their final velocity was zero. The momentum of the raindrops thus changes from \( m \times v \) to \( 0 \). The impulse, therefore, equals the initial momentum \( m \times v \), as shown in the calculation of 7874 kg\cdot m/s.
Understanding how impulse relates to momentum change is essential when calculating the effects of forces over time, especially in scenarios with non-instantaneous interactions, like rain hitting the ground.
Force calculation
When determining the force exerted by an object or required to change an object's momentum, understanding the relationship between force, impulse, and time is crucial. The force exerted can be calculated using the formula:\[ F = \frac{\text{Impulse}}{\text{time}} \]Here, \( F \) is the force, \( \text{Impulse} \) is the change in momentum (which we computed earlier), and \( \text{time} \) is the duration over which the force is applied.
This equation comes from the Impulse-Momentum Theorem, which states that the impulse on an object is equal to the change in its momentum. In our scenario with the storm, the impulse from stopping the rain over 32400 seconds led us to calculate an average force of 0.243 Newtons. This is the average force per square meter that the ground had to exert upwards to stop the rainfall.
Understanding this connection between force, impulse, and time helps to solve many real-world physics problems, from calculating braking forces to understanding how sports equipment absorbs impact.
Physical units conversion
Units conversion is a critical skill in physics problem-solving as it ensures consistency and accuracy in calculations. We often deal with various systems of measurement, such as converting inches to meters, which is essential for precise results.
In our raindrop scenario, converting the rainfall from 31 inches to meters was necessary for calculating further values like volume and force. Since 1 inch equals 0.0254 meters, the conversion to 0.7874 meters allowed us to compute the volume of rain more accurately. - Volume is calculated in cubic meters, ensuring mass in kg can be used directly, given that 1 cubic meter of water has a mass of 1000 kg. - Time conversions are vital too, like changing hours into seconds to appropriately apply formulas involving rates like force. Converting physical units is an indispensable step in scientific calculations. It aligns measured values with the units expected in equations and formulas, providing a reliable basis for physics problem solving.

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Most popular questions from this chapter

A fireworks rocket is launched vertically into the night sky with an initial speed of \(44.2 \mathrm{m} / \mathrm{s}\). The rocket coasts after being launched, then explodes and breaks into two pieces of equal mass 2.50 s later. (a) If each piece follows a trajectory that is initially at \(45.0^{\circ}\) to the vertical, what was their speed immediately after the explosion? (b) What is the velocity of the rocket's center of mass before and after the explosion? (c) What is the acceleration of the rocket's center of mass before and after the explosion?

Predict/Explain A stalactite in a cave has drops of water falling from it to the cave floor below. The drops are equally spaced in time and come in rapid succession, so that at any given moment there are many drops in midair. (a) Is the center of mass of the midair drops higher than, lower than, or equal to the halfway distance between the tip of the stalactite and the cave floor? (b) Choose the best explanation from among the following: I. The drops bunch up as they near the floor of the cave. II. The drops are equally spaced as they fall, since they are released at equal times. III. Though equally spaced in time, the drops are closer together higher up.

An \(85-k g\) lumberjack stands at one end of a \(380-\mathrm{kg}\) floating log, as shown in Figure \(9-15\). Both the log and the lumberjack are at rest initially. (a) If the lumberjack now trots toward the other end of the log with a speed of \(2.7 \mathrm{m} / \mathrm{s}\) relative to the log, what is the lumberjack's speed relative to the shore? Ignore friction between the log and the water. (b) If the mass of the log had been greater, would the lumberjack's speed relative to the shore be greater than, less than, or the same as in part (a)? Explain. (c) Check your answer to part (b) by calculating the lumberjack's speed relative to the shore for the case of a \(450-\mathrm{kg}\) log.

Object 1 has a mass \(m_{1}\) and a velocity \(\overrightarrow{\mathbf{v}}_{1}=(2.80 \mathrm{m} / \mathrm{s}) \hat{\mathrm{x}}\). Object 2 has a mass \(m_{2}\) and a velocity \(\overrightarrow{\mathbf{v}}_{2}=(3.10 \mathrm{m} / \mathrm{s}) \hat{\mathrm{y}} .\) The total momentum of these two objects has a magnitude of \(17.6 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) and points in a direction \(66.5^{\circ}\) above the positive \(x\) axis. Find \(m_{1}\) and \(m_{2}\)

A block of wood is struck by a bullet. (a) Is the block more likely to be knocked over if the bullet is metal and embeds itself in the wood, or if the bullet is rubber and bounces off the wood? (b) Choose the best explanation from among the following: I. The change in momentum when a bullet rebounds is larger than when it is brought to rest. II. The metal bullet does more damage to the block. III. since the rubber bullet bounces off, it has little effect.

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