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Q: Question [2] The truss in the figure below is supporting a 70 kN load applied at joint A and 90 kN…
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Q: * A 1- Threshold shown in Figure (2), calculate the reactions in datums A, F and 1.5 m 1.5 m 450 N…
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Q: Find the net force and its direction
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A: Given as, ρm =14kwρA =3kwρB =5kwρC =6kwN=52rps
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Q: 35. Technician A says positive caster will result from excessive toe-in. Technician B says a vehicle…
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A: The work done from state 1-2: -296.02 kJ/kg
Q: A cantilever truss shown is subjected to a wind load of 4.32 kN/m normal to the roof surface. Use $1…
A: given; load density(w)=4.32KN/mS1=0.8mS2=1.6mlets take compression force in FG member=Plets take…
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A: Given Load system To find Case 1 Horizontal force in point Case 2 Horizontal force in beam
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Q: Time is an inconsistent quantity. true Error The force generated in the truss ribs is either tension…
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Q: A steam turbine carrying a full load of 46 MW uses 77.7 kg/s of steam. The turbine efficiency is…
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Q: Threshold shown in the figure below Calculate the reactions in the datums Ϝ, A and E
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- 1) An electric motor’s torque is a function of the speed at which it rotates. Many brushed AC motors have a linear toque curve as shown below. If the power output of the motor is equal to the torque (T) multiplied by the rotational speed in rad/s (w), what is the maximum power you can get out of the motor if the stall torque (Tstall) is 0.001 Nm and the no load speed (wno load) of 10,500 RPM? Give answer in Watts.The horizontal force due to water acting on the bolt (Fh): Fh=ρ×A1×V1×V1--V2 Fh=1000×π4×0.0752×1.254×1.254+11.286( Fh)=69.472 This is the only part I did not understand how to get. Can you please explain more? I arrived with this equation, -P1A1 - P2A2 (cancelled since atmos. Pres.) +Fh = rho A1v1 (v2-v1) Why did P1A1 disappear in your solution? How did you come up with (v1 - - v2) ? What is the control volume? Why did you not do a summation of forces is equal to change in momentum but instead go immediately to getting the horizontal force acting on the bolt?URGENT.THEORY OF MACHINE A circular cutter has 4 different-sized blades, Blade A, Blade B, Blade C and Blade D, are rotating at constant velocity of 230 rad/s.The known parameters for each blade are : The weights: WA = 15 kg, WB = 25 kg, WD = 20kg The radius: RA = 300 mm, RC = 200 mm and RD = 350 mm The angle: θA = 190°; θB = 45°; θC = 90°; θD = 260° You have been asked to determine: (a) The weight of Blade C (in N) and radius of Blade B (in m) to make the whole cutter system balanced, without adding additional blade. (b) If the plane containing B and C is 0.5 m apart, determine position of planes A and D. Note: If plane A as baseline, find the length for each plane to the baseline (lb, lc, ld) (c) Sketch and label the statically balanced link of the system.
- The same engineer decides to look into rates of cooling for liquids toexperiment with different cooling solutions for servers. She finds thatthe rate of cooling for one liquid can be modelled by the equation:y = 48 × 0.99t(0 ≤ t ≤ 80)where y is the temperature of the liquid in degrees Celsius and t is thetime in minutes.(i) State whether the type of reduction for this model is linear orexponential. Describe how reduction rate differs between linear andexponential functions. (ii) Calculate the temperature when t = 20. [3](iii) Write down the scale factor and use this to find the percentagedecrease in the temperature per minute. (iv) Use the method shown in Subsection 5.2 of Unit 13 to find the timeat which the temperature is 30◦(v) Determine the halving time of the temperature.Example 9.15 A rope, to which a weight is attached, passes around a pulley 50 cm in diameter. The angular acceleration of the pulley is 18 rad/s2. If the pulley is initially rest, find a- The time required for the weight to attain a velocity of 15 m/s b- The number of revolution through which the pulley rotates during that period, c- The total acceleration of a point on the rim of the pulley 0.5 second after it was at rest.Match the laws accordingly. 1.1st law of thermodynamics 2.2nd law of thermodynamics (foward engine) 3.Joules law 4.2nd law of thermodynamics (reverse engine) 5.3rd law of thermodynamics 6.second law of thermodynamics (Carnot) A. The internal energy of a perfect gas is a function of the absolute temperature only B. No Heat Engine can be more efficient than a Reversible Heat Engine operating between the same temperature limits C. A pure crystalline substance at absolute zero temperature is in perfect order, and its entropy is zero D. When a system undergoes a complete cycle, the net Heat supplied plus the net Work input is zero. E. It is impossible for a heat engine to produce a net work output in a complete cycle if it exchanges heat only with a single energy reservoir. F. It is impossible to construct a device that operating in a cycle will produce no effect other than the transfer of heat from a cooler to a hotter body.
- Complete the following table for refrigerant-134a. Show your work and explain how you got your values. T (oF) P psia) h (Btu/Ibm) X Phase description 80 78 15 0.6 10 70 180 129.46 110 1.0The equations below are the horizontal xx- and vertical yy-component forms of Newton's second law applied to a physical process.(5.0kg)ax=(50N)cos30∘+Ncos90∘−0.5Ncos0∘+(5.0kg)(9.8N/kg)cos90∘(5.0kg)ax=(50N)cos30∘+Ncos90∘−0.5Ncos0∘+(5.0kg)(9.8N/kg)cos90∘(5.0kg)0=(−50N)sin30∘+Nsin90∘+0.5Nsin0∘−(5.0kg)(9.8N/kg)sin90∘ Determine N Determine ax Invent a problem for which the equations might provide an answer.According to the U.S. Geological survey, the volume of water in oceans, seas, and bays is 1.41 ✕ 1018 m3. The average concentration of naturally occurring uranium in seawater is 3.15 ✕ 10−6 kg/m3. If 0.670% of naturally occurring uranium is the fissionable isotope U-235 and each fission reaction yields 200 MeV, estimate the amount of time the uranium in the seawater on the planet could meet the planet's energy needs at an average usage of 1.94 ✕ 1013 J/s.
- What is the change in internal energy (ΔU) for these processes? Remember that ΔU = (3/2)nRΔT = (3/2)NΔT Isobaric:P = +20.00V = +48.33T = +48.33N = +20.00 Isochoric:P = +20.00V = +100.00T = +49.00N = +20.00 Adiabatic:P = +3.50V = +284.06T = +49.67N = +20.00In the design of a jet engine part, the designer has a choice ofspecifying either an aluminium alloy casting or a steel casting. Either materialwill provide equal service, but the aluminium casting will weigh 1.2 kg ascompared with 1.35 kg for the steel casting.The aluminium can be cast for $ 80.00 per kg. and the steel one for$ 35.00 per kg. The cost of machining per unit is $ 150.00 for aluminiumand $ 170.00 for steel. Every kilogram of excess weight is associated witha penalty of $ 1,300 due to increased fuel consumption. Which materialshould be specified and what is the economic advantage of the selection perunit?....A steam engine develops power at mean torque of 19,000 N-m and velocity of 10 m/s. The mean diameter of the flywheel rim is 1.2 meters. The coefficient of fluctuation of speed is 0.03 and the coefficient of fluctuation of energy is 0.21. Determine the thickness of the rim in mm if the width of the rim is twice of its thickness. Assume density of rim material as 7250 kg/m3.