9514 1404 393
Answer:
746.7 ft²
Explanation:
You can add them up, or you can take advantage of multiplication to make the repeated addition simpler.
(112.5 ft² +136.4 ft²) +(112.5 ft² +136.4 ft²) +(112.5 ft² +136.4 ft²)
= (3)((112.5 ft² +136.4 ft²) = 3(248.9 ft²) = 746.7 ft²
The total area of the decks on the 3 homes is 746.7 ft².
Answer:
Volume of aeration tank = 1.29 x 10^4 m³
Explanation:
Food/Micro- organism Ratio = 0.2/day
Feed Rate (Q) = 0.438 m³/s
Influent BOD = 150 mg/L
MLVSS = 2200 mg/L
The above mentioned parameters are related by the equation
F/M = QS₀/VX
where S₀ is the influent BOD and X is the cell concentration, with V being the Volume of the tank. Re- arranging the above equation for V and putting the values in, we get
V = 0.4380 x 150/0.2 x 2200
V = 0.1493 (m³/s) x day
V = 0.1493 x 24 x 60 x 60
V = 1.29 x 10^4 m³
Answer:
Explanation:
The concept of Hooke's law was applied as it relates to deformation.
The detailed steps and appropriate substitution is as shown in the attached file.
Answer:
5.7058kj/mole
Explanation:
Please see attachment for step by step guide
Answer:
The problem is that the pumps would consume more energy than the generators would produce.
Explanation:
Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.
A pump uses electricity to add energy to the water to send it to a higher potential energy state.
Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.
What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.
Answer:
(a) ΔU = 125 kJ
(b) ΔU = -110 kJ
Explanation:
(a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process?
The work is done to the system so w = 150 kJ.
The heat is released by the system so q = -25 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -25 kJ + 150 kJ = 125 kJ
(b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?
The work is done by the system so w = -100 kJ.
The heat is released by the system so q = -10 kJ.
The change in internal energy (ΔU) is:
ΔU = q + w
ΔU = -10 kJ - 100 kJ = -110 kJ
What type of search do you think this would be?
This type of search would be a linear search. A linear search involves looking at each item in the array one by one until the desired item is found.
In this case, you would look at each square in the array until you find the one that matches the object pictured on the left. This is a common search method for small arrays or when the array is unsorted.
The algorithm continues until the target element is found, or until it reaches the end of the array and fails to find the target element. Linear searches are useful in scenarios where the array is small and unsorted, as the algorithm does not need to compare every element in the array to find the target element.
Learn more about linear search:
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