Tutor profile: Edmund T.
It's almost summer time, and the local field club is about to hire a company to fill its pool full of water. The pool company charges by the hour and by the gallon of water added to the pool. The club's pool is 4 ft deep throughout the entire pool, 25 yards long, and 15 yards wide. If the company charges $0.10 per gallon and $120 per hour and fills the pool at a rate of 10 gallons per minute, how much will filling the pool cost the field club? Assume 7.48 cubic feet is equal to 1 gallon.
We can immediately can see that we're going to have to determine the volume of the pool in gallons, but we only have its dimensions given in feet and yards, so we will have to do some unit conversion. Let's first find our pool dimensions in feet. Knowing that 1 yard = 3 feet, we can easily determine that the pool is 75 ft long, 45 ft wide, and 4 ft deep. To determine the volume of the pool in cubic feet, we multiply the product of 75 and 45 by 4. This gives a volume of 13,500 cubic feet. Since we know that 7.48 cubic feet is equal to 1 gallon, we can convert 13,500 cubic feet to gallons by dividing by 7.48. This gives a volume in gallons of 1804.8 gallons. We now know how much the company will charge for the volume of water added, but we also need to know how many hours the company will spend filling the pool. We know that the company fills the pool at a rate of 10 gallons per minute. To make this more useful, we need to convert this to gallons per hour, which we can do by multiplying by 60. This gives us a fill rate of 600 gallons per hour. Now, to determine how many hours it will take to fill the pool, we divide the volume of the pool by the gallon per hour fill rate: 1804.8 divided by 600, which equals 3.01 hours. To determine how much the company will charge for the hours spent tilling the pool, we multiply $120 per hour by 3.01 hours, which gives us $361.20. To determine how much the company will charge for the volume of water used, we multiply $0.10 per gallon times 1804.8 gallons, which gives us $180.48. Adding the two expenses, $361.20 and $180.48, we solve that the total expense the field club will have to pay the pool company is $541.68.
You and your dad want to move your new laundry machine from the warehouse back to your house. You plan on using your dad's truck, but you are worried that the laundry machine will tip over in the truck bed while you drive it home. What driving instructions can you give your dad to prevent your new laundry machine from tipping over? The square-shaped laundry machine is 25 kg and 1 m tall. Assume that gravity is equal to 10.0 m/s^2; the center of mass of the machine is its geometric center; this is a 2-D problem that considers one of the machine's bottom edges as a point of rotation; and that the friction between the bed and machine will prevent it from slipping before tipping over.
This is a wordy problem, and we are given hardly any numbers to work with. What are we going to do, then? We start with what we know! For starters, we know that gravity is 10.0 m/s^2 and acts on the center of mass of the machine. We also know that the machine weighs 25 kg. Therefore, from that information alone, we know how much gravitational force is acting on the machine, and it is acting in the downward direction. Assuming that the bottom edges of the machine are its points of rotation, we can think of gravity as the moment that keeps the machine from rotating about any point. Therefore, for the machine to tip over or rotate, some force will have to overcome that force of gravity. In this problem's case, that force is the relative force on the machine caused by its mass, or inertia, and the acceleration of the truck. This inertial force can be represented as the mass of the machine times the acceleration of the truck, but in the opposite direction of the truck's acceleration: F = m*a. The force of gravity can be represented as the mass of the machine times the acceleration of gravity in the downward direction: Fg = m*g. Both of these forces act perpendicularly and the same distance from the point of rotation because they both act at the center of mass. Therefore, we do not have to worry about the length of the moment arm of rotation, and, thus, we know the machine will tip simply when the acceleration of the truck is greater than the acceleration due to gravity. So tell your dad that he can drive at whatever speed he wants, but he can't accelerate more than 10 m/s per second!
A customer has $2.11 worth of change in her wallet. She knows that she has three times as many quarters as she does dimes, one-fourth times as many dimes as she does nickels, and eight times as many nickels as she does pennies. How many of each coin does the customer have?
This problem can be difficult because instead of giving us the numbers and equations to solve it, we have to create them ourselves. But one thing we do know is that the customer's change will add up to $2.11. We also know the value of each coin: $.25 for a quarter; $.10 for a dime; $0.05 for a nickel; and $0.01 for a penny. Lastly, we also know how many of these coins we have relative to each other, and we can create equations based on this information. Let's make variables for the number of each coin: Q for quarters; D for dimes; N for nickels; and P for pennies. Now we can create our equations: Q = 3D D = 1/4N N = 8P Since we know that the sum of the values of each coin times the numbers of each coin will equal the total change, we can create another equation: 0.25Q + 0.10D + 0.05N + 0.01P = 2.11 Now that we have four equations, we can solve for our four unknown variables, Q, D, N, and P, by using substitution. Since we know that Q = 3D, let's substitute 3D for Q 0.25(3D) + 0.1D + 0.05N + 0.01P = 2.11 0.75D + 0.1D + 0.05N + 0.01P = 2.11 0.85D + 0.05N +0.01P = 2.11 Now, let's substitute 1/4N for D: 0.85(1/4N) + 0.05N + 0.01P = 2.11 0.85(.25N) + 0.05N + 0.01P = 2.11 0.2125N + 0.05N + 0.01P = 2.11 0.2625N + 0.01P = 2.11 And now, substitute 8P for N: 0.2625(8P) + 0.01P = 2.11 2.1P + 0.01P = 2.11 2.11P = 2.11 Now, solve for P P = 2.11/2.11 P = 1 The customer has 1 penny! Now we can solve our other equations: N = 8P N = 8(1) N = 8 D = 1/4N D = 1/4(8) D = 2 Q = 3D Q = 3(2) Q = 6 Therefore, based on our calculations, the customer has 6 quarters, 2 dimes, 8 nickels, and 1 penny. Let's check our math: 6(0.25) + 2(0.10) + 8(0.05) + 1(0.01) = 2.11 1.50 + 0.20 + 0.40 + 0.01 = 2.11 Since we checked our calculations, we can be sure that the customer indeed has 6 quarters, 2 dimes, 8 nickels, and 1 penny. Awesome! We solved our problem.
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