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A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force Fc1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to Fc2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)–0.4, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force Fc2. i.e.,x100, is _______ (round off to one decimal place).Correct answer is '-5.6'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force Fc1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to Fc2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)–0.4, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force Fc2. i.e.,x100, is _______ (round off to one decimal place).Correct answer is '-5.6'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force Fc1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to Fc2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)–0.4, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force Fc2. i.e.,x100, is _______ (round off to one decimal place).Correct answer is '-5.6'. Can you explain this answer?.

Solutions for A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force Fc1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to Fc2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)–0.4, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force Fc2. i.e.,x100, is _______ (round off to one decimal place).Correct answer is '-5.6'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE. Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.

Here you can find the meaning of A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force Fc1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to Fc2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)–0.4, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force Fc2. i.e.,x100, is _______ (round off to one decimal place).Correct answer is '-5.6'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force Fc1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to Fc2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)–0.4, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force Fc2. i.e.,x100, is _______ (round off to one decimal place).Correct answer is '-5.6'. Can you explain this answer?, a detailed solution for A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force Fc1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to Fc2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)–0.4, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force Fc2. i.e.,x100, is _______ (round off to one decimal place).Correct answer is '-5.6'. Can you explain this answer? has been provided alongside types of A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force Fc1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to Fc2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)–0.4, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force Fc2. i.e.,x100, is _______ (round off to one decimal place).Correct answer is '-5.6'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A cylindrical bar with 200 mm diameter is being turned with a tool having geometry 0° - 9° - 7° - 8° - 15° - 30° - 0.05 inch (Coordinate system, ASA) resulting in a cutting force Fc1. If the tool geometry is changed to 0° - 9° - 7° - 8° - 15° - 0° - 0.05 inch (Coordinate system. ASA) and all other parameters remain unchanged, the cutting force changes to Fc2. Specific cutting energy (in J/mm3) is Uc = U0 (t1)–0.4, where U0 is the specific energy coefficient, and t1 is the uncut thickness in mm. The value of percentage change in cutting force Fc2. i.e.,x100, is _______ (round off to one decimal place).Correct answer is '-5.6'. Can you explain this answer? tests, examples and also practice GATE tests.