The composite C value for the given drainage area for a 10-year storm recurrence interval is 0.45. Solution: Calculate composite C by using the following equation: C = (C1A1 + C2A2 + C3A3) / (A1 + A2 + A3) C = [(0.25 acres x 0.49) + (0.25 acres x 0.22) + (0.10 acres x 0.95)]/ (0.25+0.25+0.10) C = 0.45
Answer: Na+ Electrolytes conduct electricity in aqueous solution, and are typically found in the form of ions. In the context of the question, Na+ is the only electrolyte.
Air-entrainment was a significant concrete technology advancement made in the 1930s to provide relief from internal pressure. Entrained air provides tiny areas within concrete into which water can expand when it freezes. Air-entrained concrete is recommended when a concrete structure is exposed to freezing conditions and is achieved by adding surfactants to a concrete mix.
The hydraulic radius of a sewer refers to the ratio of the cross-sectional area of flow to the wetted perimeter. (The wetted perimeter is the portion of a cross-section’s perimeter that is “wet.”) The equation that describes the hydraulic radius of a channel, Rh, is expressed as follows: Rh = A (cross sectional area of flow) / P (wetted perimeter)
Answer: Sodium Solve this question with your knowledge of the periodic table. The most reactive family of metals is the alkali metals. In the context of the question, only one metal is part of this family, sodium. Thus, sodium will react explosively when placed in water.
Answer: Na2O Metallic oxides consist of an oxygen atom bound to a metal. The only metal present in this question is sodium, so Na2O must be a metallic oxide.
- Fresh concrete mix is sampled and tested by an inspector at a construction site. Field inspectors are typically required to do the following: · check the mix design and obtain representative samples of reinforcing steel, joint materials, and curing compounds · sample fresh concrete · test fresh concrete for slump, air content, and temperature
Answer: C2H4 Hydrocarbons with the formula CnH2n contain double bonds. The only compound with this formula, C2H4, must contain a double bond.
Answer: Lithium Easily oxidized elements readily release electrons. The loss of an electron allows these elements to form a stable valence electron configuration. In the context of this question, lithium is the most willing to release an electron, so it is the most easily oxidized.
The porosity of a 950 g dry soil sample, with an undisturbed volume of 900 cm3 and a particle density of 2.65 g/cm3, is 60%. The porosity of a soil is the ratio of the volume of soil pores to the total soil volume, and the soil porosity values range from 0 to 1. Solution: Use the following equation to determine porosity, n n = 1 - bulk density / particle density = 1 - Db / Dp Given Dp = 2.65 Therefore, n = 1 - [(Mass of dry soil / volume of solids and pore spaces)/ 2.65] = 1 - [(950 g) / (900 cm3)] / 2.65 g/cm3 x 100 n = 60%
Answer: Copper The delocalized “sea of electrons” model describes bonding in metals. Therefore, the only metal, copper, must display this type of bonding.
The Rankine and Coulomb Theories are used to analyze active and passive lateral earth pressure. The two theories are similar; however, the Rankine Theory assumes there is no adhesion or friction between the wall and the soil, and lateral pressure is limited to vertical walls. The Coulomb Theory assumes that there is friction between the wall and the soil and that lateral pressure is not limited to vertical walls.
Water hammer is the impulsive force of water that is stopped too quickly. In general, the maximum recommended average velocity in a closed pipeline is 5 feet/second. When the velocity of flow within a pipeline exceeds 5 feet/second, pressure surges may develop within the pipeline when flow is stopped suddenly. The rapid change in momentum of the water causes this impulsive force.
The passing sight distance for a section of 60 mph 2-lane rural highway with a 1,740-foot vertical sag curve where a bridge passes over at 16.8 feet is 2,273 feet. Solution: Use sag vertical curve length at undercrossing equation (US Customary units) S = L/2 + 400(C-5.75)/A Where, L = 1740 feet C = 16.8 A = 3.15 S = 1740/2 + 400(16.8-5.75)/3.15 = 2,273 feet
An open-pit excavation would not be permitted to construct a spread footing in cases where the soil moisture content is variable. Open-pit excavation is only recommended when soils can be maintained in a dry condition. Shoring and cribbing is needed in shallow foundation excavations where soil moisture content can vary widely due to the influence of groundwater or surface water runoff.
An appropriate checklist for rebar inspection is recommended that includes the listed items, independent testing laboratory reports, and approved placing drawings. CRSI recommends that a checklist procedure and the minimum requirements for rebar acceptance should be established during the pre-construction conference. These quality control measures will help to ensure that the rebar used will meet specific material standards and the project owner’s requirements.
The statement, “Sizing is based on the measured flow conditions” is not correct when considering the surface overflow rates of primary settling tanks. Typically, the primary settling tanks are sized based on the surface area calculations performed using 1) the design average flow conditions and 2) design peak hourly flow conditions. The larger surface area of the two is used to size the primary setting tank. Specific design requirements for primary settling tanks are set by state regulating agencies.
Answer: 7 Begin this problem by considering the oxidation state of oxygen. As in most oxides, oxygen has an oxidation state of -2. Multiply this value by four to account for all of the oxygens, giving a total oxidation value of -8. Potassium has an oxidation state of +1, bringing the total oxidation value to -7. In order to maintain electrical neutrality, chlorine must have an oxidation state of +7.
Footings are appropriate to use when foundation loads can be supported by shallow soils. Reinforced concrete footings are designed to handle the foundation loads and local soil characteristics. Deep foundations may be used when shallow soils are not able to support the foundation loads cost-effectively. Piles may be used when deep, competent soil or bedrock can be used to support the foundation loads.
The horizontal distance H for a slope measurement of 814.20 ft and angle of 6°30’ is 808.96 ft. Solution Use the horizontal distance formula H = L cos a Where L = 814.20 ft A = 6°30’ Step 1 Convert angle 6°30’ to DMS: 6°30’ = 6.5° Step 2 H = L cos 6.5 = (814.20 ft)(0.994) = 808.96 ft
