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Chapter 7: Problem 30

In a severe storm, \(1.00 \mathrm{~cm}\) of rain falls on a flat horizontal roof in \(30.0 \mathrm{~min}\). If the area of the roof is \(100 \mathrm{~m}^{2}\) and the terminal velocity of the rain is \(5.00 \mathrm{~m} / \mathrm{s}\), what is the average force exerted on the roof by the rain during the storm?

Short Answer

Expert verified
Answer: The average force exerted on the roof by the rain during the storm is 2.78 N downward.

Step by step solution

01

Convert the amount of rain to meters

First, we need to convert the amount of rain from centimeters to meters. We are given that 1.00 cm of rain falls during the storm. We know that 1 cm = 0.01 m. So, we can convert it as: $$ 1.00 \mathrm{~cm} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.0100 \mathrm{~m} $$
02

Determine the volume of rain fallen on the roof

Next, we need to determine the volume of rain falling onto the roof during the storm. We can find this by multiplying the area of the roof by the amount of rain that falls: $$ \mathrm{Volume} = \mathrm{Area} \times \mathrm{Height} $$ $$ \mathrm{Volume} = 100 \mathrm{~m^2} \times 0.0100 \mathrm{~m} = 1.00 \mathrm{~m^3} $$
03

Calculate the mass of the rain fallen

To find the mass of the rain, we need to consider the density of water, which is approximately \(1000 \mathrm{~kg/m^3}\). We can use the density formula to find the mass of the rain: $$ \mathrm{Density} = \frac{\mathrm{Mass}}{\mathrm{Volume}} $$ $$ 1000 \mathrm{~kg/m^3} = \frac{\mathrm{Mass}}{1.00 \mathrm{~m^3}} $$ Solving for Mass, $$ \mathrm{Mass} = 1000 \mathrm{~kg/m^3} \times 1.00 \mathrm{~m^3} = 1000 \mathrm{~kg} $$
04

Calculate the time interval of the rain

To calculate the average force exerted, we need to know the time interval of the rain in seconds. The problem states that the rain lasts 30.0 minutes, equivalent to 1800 seconds.
05

Calculate the momentum change of the rain

As the rain falls onto the roof, its velocity goes from the terminal velocity to 0. Therefore, the change in momentum can be found using the relationship: $$ \Delta P = m \times \Delta v $$ $$ \Delta P = 1000 \mathrm{~kg} \times (0 \mathrm{~m/s} - 5.00 \mathrm{~m/s}) $$ $$ \Delta P = -5000 \mathrm{~kg \cdot m/s} $$ The negative sign indicates that the momentum is directed towards the roof.
06

Calculate the average force exerted by the rain

Finally, we can calculate the average force exerted on the roof using the relationship: $$ F_{avg} = \frac{\Delta P}{\Delta t} $$ $$ F_{avg} = \frac{-5000 \mathrm{~kg \cdot m/s}}{1800 \mathrm{~s}} $$ $$ F_{avg} = -2.78 \mathrm{~N} $$ The negative sign indicates the force is directed downwards, which is the force exerted on the roof by the rain during the storm. So, the average force exerted on the roof by the rain during the storm is 2.78 N downward.

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Most popular questions from this chapter

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