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We go out to sunbathe on a warm summer day. If we soak up 100 British thermal units per hour [BTU/h] of energy, how much will the temperature of 132 pound-mass [lbm] person increase in 2 hours [h] in units of degrees Celsius [°C]? We assume that since our bodies are mostly water they have the same specific heat as water, 4.18 joules per gram degree Celsius [J/(g °C)].
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INTERNATIONAL EDITION---Engineering Mechanics: Statics, 14th edition (SI unit)
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- #01: Explain the following terms in your own words. a) Aerodynamics. b) Mach Number. c) Sub-Sonic. d) Angle of Attack. e) Thrust. f) Chord Line. g) Drag Coefficient. h) Pitch Moment. i) Wind Tunnel. j) Skin Drag.arrow_forwardThe amount of time it takes for a pendulum to swing “to-and-fro” through one complete cycle is called the “period” and is calculated using the following equation: where T = period in seconds, I = mass moment of inertia, m = mass of pendulum in kilograms g = gravitational acceleration, 9.8 m/s2, h = effective length of pendulum in meters What is the appropriate unit for I if the preceding equation is to be homogeneous in units? Please show all of your work.arrow_forwardCompletely solve and box the final answer. Please show the units to show the unit cancellations. Write legibly 2. During a steady flow process, the pressure of the working substance drops from 1,380 kpa to 138 kpa, the speed increases from 61 m/s to 305 m/s, the internal energy of the open system decreases 58.1 KJ/kg, and the specific volume increases from 0.0625 m3/kg to 0.5 m3/kg. No heat is transferred, determine the work in KW if the mass flow at the rate of 272.15 kg/hr.arrow_forward
- Process Fluid Mechanics for Chemical Engineers , please help in this Question , thanks.arrow_forward1. Energy consumption in a country from 2001 to 2011 increased from 1041.3 million to 1526.1 million BOE (barrels of oil equivalent). If the high calorific value (HHV) of the oil is 141,000 kJ/gallon what is the increase in energy consumption over 10 years in Joule units. What is the rate of energy consumption per year? 2. Based on data on the rate of energy consumption per year from question no. 1, if the growth rate energy 2.5% per year, calculate the doubling time (td) and what is the approximate rate of energy growth in 2021.arrow_forwardSuppose you're converting a bill of materials for machining feedstock from the British gravitational (U.S. system) of units to International System (SI) units. Your goal is to have all of the material quantities in kilograms. Your bill of materials lists values for required weights of brass rod in pounds. To convert to these values to kilograms you first you divide by gravitational acceleration in feet per second squared to get a mass in slugs, then convert slugs to kilograms using standard conversion factors. Group of answer choices True Falsearrow_forward
- Seawater can be purified by distillation into potable water. If it costs 15 cents per kilowatt-hour for energy, what is the cost of energy to produce 100.0 gallons of drinking water at 100.0 ˚C if the seawater starts at 25.00 ˚C? (1 watt × 1 second = 1 joule). Enter your answer to the nearest penny.arrow_forwardFor the Following question Graph all 4 : [I just need all 4 graphs and please explain and make clean solution] Position vs time Velocity vs time Acceleration vs time Force vs time [For your convenience, I have solved the numerical solutions for the problem] (Please Look at the picture since it is much cleaner) Question : A 550 kilogram mass initially at rest acted upon by a force of F(t) = 50et Newtons. What are the acceleration, speed, and displacement of the mass at t = 4 second ? a =(50 e^t)/(550 ) [N/kg] v = ∫_0^t▒(50 e^t )dt/(550 )= v_0 +(50 e^t-50)/550=((e^t- 1))/11 x = ∫_0^t▒(e^t- 1)dt/(11 )= x_0 +(e^t- t - 1)/(11 ) a(4s)=(50*54.6)/550= 4.96[m/s^2 ] v(4s)=((e^4-1))/11= 4.87[m/s] x(4s)=((e^4- 4 - 1))/11= 4.51 [m]arrow_forwardTires are one of the most frequently encountered applications of the gas laws that we never think about. We fill our tires with air, or with nitrogen, but it always works out the same way. Enough gas goes in, the tire inflates, and then the pressure starts going up. In this assignment, we’ll be investigating the ways that the gas laws impact how we treat our tires. Q1. I have good information that in Fast 29, Dominic Toretto (Vin Diesel) will need to refill a tire quickly during a dramatic moment. For this reason, he has a 3.00 L tank of compressed air that is under 2.7892*103 mmHg and is kept cool in dry ice at -35.0 °C. When Dom hooks his compressed air up to his completely empty 10.50 L tire at 39.2 °C and lets it run, what will his final tire pressure be, in atm? Assume all the air is transferred into the tire. Is his tire pressure above the 2.31 atm that he needs to save the planet/his family/his crew? Q2. Your car tire pressure sensor looks to see when your tire pressure…arrow_forward
- The Ideal gas law shows the relationship between some common properties of ideal gases and is written as PV = nRT. P = pressure, V = volume, n = number of moles of ideal Gas, R = General gas constant = 8.314 kJ/(kmol.K) and T = absolute temperature. 5 mol ideal gas at 22°C is placed in a cube with edge lengths of 0.750 meters. Calculate the pressure of the Ideal gas as [Pa].arrow_forwardExperiment #2: Acceleration vs. Mass Link https://phet.colorado.edu/sims/html/forces-and-motion-basics/latest/forces-and-motion-basics_en.html In this lab you will determine the relationship between acceleration and mass. Choose an Applied Force at the beginning, and keep it constant for this entire experiment. Set the friction to zero. This will make your Applied Force equal to the net force. Record data for five different values of Mass. Graph Acceleration vs. Mass. Graph this in Google sheets(you want a line graph - it should only have one line). Make sure that Mass information is used as the x value Make sure that Acceleration information is used as the y value Add a trendline – see what fits best – linear, exponential, polynomial, etc … Add a copy of you graph below the table Applied Force (N) Mass (kg) Acceleration (m/s²)arrow_forwardThe differential change in pressure p close to the surface of a static fluid is given by the following expression:dp/dy = -3Ap2,where A is a constant, with units of 1/(atm•m), and p is the pressure in atm. The pressure at the surface of the fluid is p(0) = 1 atm, and the coordinate y here is positive upwards with origin at the surface. An absulute pressure gauge is placed at a depth 0.19m in the fluid. What would be the reading of the pressure gauge in units in atm? you can take the constant A=1(atm.m)^-1arrow_forward
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