9. Power is transferred between an electric motor and a wheel via a chain drive. The motor and wheel shafts are 19.7 in apart and the wheel shaft must rotate 1200 rpm. The driving sprocket is installed directly on the motor shaft and has 23 teeth. The motor rotates at a speed of 1500 rpm. The chain has a pitch of 0.5 in. Calculate the number of teeth of the driven sprocket and chain length. option a. Driven sprocket has 29 teeth and chain length is 106 pitches. b. Driven sprocket has 19 teeth and chain length is 106 pitches. command c. Driven sprocket has 19 teeth and chain length is 101 pitches. d. Driven sprocket has 29 teeth and chain length is 101 pitches. e. Impossible to tell from the given data.
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- You have a 3-phase motor, M1, which drives a conveyor belt. The conveyor belt is operated with switches S1 (Start- forward), S2 (Start- back) and S0 (Stop). S1 and S2 are pressure switches with work contact (NO), SO is pressure switch with rest contact (NC). All switches have automatic return. The operator must operate the switches with one push to start the conveyor belt, it should not be necessary to hold the switches in for the conveyor belt to start. In order to change direction, the operator must first press stop, it will then take 5 seconds before it is possible to start the belt again. a. Draw the main flow diagram. Remember that the motor must have a direction of rotation reverser b. Draw control flow diagramTwo masses are connected to a string and hung on a pulley to make an atwood machine. The masses are different values such as M > m. The setup is shown. Assume the string is massless and the pulley is ideal. When the masses are gently released, there is a resulting acceleration. Which of the following equations will calculate the mass of m in the system if the big mass and acceleration are known? A) m = 2M(a/g). B) = m 2M(g/a). C) m = M(g- a/g+a). D) m = M(g+a/g -a).Life Force Fitness, Inc., assembles and sells treadmills. Activity-based product information for each treadmill is as follows: Activity Activity-BaseUsage(hrs. per unit) X Activity Rateper Hour = ActivityCost Motor assembly 1.50 $20 $30.00 Final assembly 1.00 18 18.00 Testing 0.25 22 5.50 Rework 0.40 22 8.80 Moving 0.20 15 3.00 Activity cost per unit $65.30 All of the activity costs are related to labor. Management must remove $2.00 of activity cost from the product in order to remain competitive. Rework involves disassembling and repairing a unit that fails testing. Not all units require rework, but the average is 0.40 hours per unit. Presently, the testing is done on the completed assembly; but much of the rework has been related to motors, which can be tested independently prior to adding the motor to…
- please answer within 30 minutes.Life Force Fitness, Inc., assembles and sells treadmills. Activity-based product information for each treadmill is as follows: Activity Activity-Base Usage(hrs. per unit) X Activity Rateper Hour = Activity Cost Motor assembly 1.50 $20 $30.00 Final assembly 1.00 18 18.00 Testing 0.25 22 5.50 Rework 0.40 22 8.80 Moving 0.20 15 3.00 Activity cost per unit $65.30 All of the activity costs are related to labor. Management must remove $2.00 of activity cost from the product in order to remain competitive. Rework involves disassembling and repairing a unit that fails testing. Not all units require rework, but the average is 0.40 hours per unit. Presently, the testing is done on the completed assembly; but much of the rework has been related to motors, which can be tested independently prior to adding the motor…Life Force Fitness, Inc., assembles and sells treadmills. Activity-based product information for each treadmill is as follows: ETT Activity-Base Usage (hrs. per unit) Activity Rate per Hour Activity Motor assembly Activity Cost $30.00 1.50 $20 Final assembly Testing 1.00 18 18.00 0.25 22 5.50 Rework 0.40 22 8.80 Moving Activity cost per unit 3.00 $65.30 0.20 15 All of the activity costs are related to labor. Management must remove $2.00 of activity cost from the product in order to remain competitive. Rework involves disassembling and repairing a unit that fails testing. Not all units require rework, but the average is 0.40 hour per unit. Presently, the testing is done on the completed assembly; but much of the rework has been related to motors, which can be tested independently prior to adding the motor to the treadmill during final assembly. Thus, motor issues can be diagnosed and solved without having to disassemble the complete treadmill. This change will reduce the average rework…
- Cycle USA is in the business of making innertubes for bicycles. They manufacture both large and small innertubes. Both of the tubes use the same material but require different amounts of material. The material usage is as follows: Large Small Rubber 3 feet at $0.25 per foot 1.25 feet at $0.25 per foot Connector 1 at $0.03 1 at $0.03 On June 1, Cycle USA purchased 25,000 feet of rubber for $6,875. During the month of June, Cycle USA used 14,500 feet of rubber to make 3,000 large tubes and 4,000 small tubes. A. Calculate the direct materials price variance, the direct materials quantity variance, and the total direct materials cost variance? Direct Materials Price Variance Direct Material Quantity Variance Total Direct Materials Cost Variance B. If Cycle USA bought 10,000 connectors costing $310, what would the direct materials price variance be for the connectors? Direct Materials Price…Shown at right are two boxes (mass m and 2m) attached to identical springs, and attached to each other by a string running over a pulley. The springs are initially neither stretched nor compressed and the masses are initially at rest. The blocks then speed up and eventually slow down again until each block has moved a distance s, at which point both blocks are momentarily at rest. Determine a symbolic expression in terms of given variables for the spring constant of each identical spring. No explanation necessary. We'll use the work-energy theorem with a system consisting of both blocks, the string, both springs, and the Earth. The net external work on this system is zero (the various normal forces all act on points of contact that are not moving). The initial and final kinetic energy for both blocks is zero because they are not moving 2m m ΔΕ = 0 ks² · − 0 + ¼½ks² − 0 + mgs − 0 + (−2mgs) − 0 = 0 k = mg SEe 135.
- Use the following information for Exercises 5-44 through 5-46: The following six situations at Diviney Manufacturing Inc. are independent. a. A manual insertion process takes 30 minutes and 8 pounds of material to produce a product. Automating the insertion process requires 15 minutes of machine time and 7.5 pounds of material. The cost per labor hour is 12, the cost per machine hour is 8, and the cost per pound of materials is 10. b. With its original design, a gear requires 8 hours of setup time. By redesigning the gear so that the number of different grooves needed is reduced by 50%, the setup time is reduced by 75%. The cost per setup hour is 50. c. A product currently requires 6 moves. By redesigning the manufacturing layout, the number of moves can be reduced from 6 to 0. The cost per move is 20. d. Inspection time for a plant is 16,000 hours per year. The cost of inspection consists of salaries of 8 inspectors, totaling 320,000. Inspection also uses supplies costing 5 per inspection hour. The company eliminated most defective components by eliminating low-quality suppliers. The number of production errors was reduced dramatically by installing a system of statistical process control. Further quality improvements were realized by redesigning the products, making them easier to manufacture. The net effect was to achieve a close to zero-defect state and eliminate the need for any inspection activity. e. Each unit of a product requires 6 components. The average number of components is 6.5 due to component failure, requiring rework and extra components. Developing relations with the right suppliers and increasing the quality of the purchased component can reduce the average number of components to 6 components per unit. The cost per component is 500. f. A plant produces 100 different electronic products. Each product requires an average of 8 components that are purchased externally. The components are different for each part. By redesigning the products, it is possible to produce the 100 products so that they all have 4 components in common. This will reduce the demand for purchasing, receiving, and paying bills. Estimated savings from the reduced demand are 900,000 per year. 5-45 Driver Analysis Refer to the information for Diviney Manufacturing on the previous page. Required: CONCEPTUAL CONNECTION For each situation, identify the possible root cause(s) of the activity cost (such as plant layout, process design, and product design).Please do not provide answer in image formate, thank you PLEASE INCLUDE CALCULATIONS!!A cement grinding mill "X" with a capacity of 48.072 tons per hour utilizes forged steel grinding balls costing P12,474 per ton, which have a wear rate of 105 grams per ton cement milled. Another cement mill "Y" of the same capacity uses high chrome steel grinding balls costing P62,000 per ton with a wear rate of 30 grams per ton cement milled. Determine the more economical grinding mill, considering other factors to be the same. What is the total cost for grinding mill "X"? (amount in PhP/hr)