D. = 3.710cm D. = 6.251cm D. = 14.90cm If an effort of 4.31N is needed to lift a load of 40.03N. determine the following
Posted: Sun May 15, 2022 11:24 pm
D. = 3.710cm D. = 6.251cm D. = 14.90cm If an effort of 4.31N is needed to lift a load of 40.03N. determine the following: a) The velocity ratio (VR). Answer: Question 2 Not yet answered Marked out of 1.00 Flag question b) The mechanical advantage Answer: Question 3 Not yet answered Marked out of 1.00 P Flag question c) The efficiency. Answer in % Answer:
d) The ideal effort (P). Answer in N. Answer: Question 5 Not yet answered Marked out of 1.00 Flag question A 40 kg block of aluminum is hung from the end of a vertical 0.50 m long steel wire with a cross-sectional area of 5.0 x 105m? Young's modulus of aluminum: 7.0 x 1010 Pa Young's modulus of steel: 2.0 x 104 Pa Tensile strength of steel: 3.0 x 10* Pa Using the information above, determine the following: a) The stress in the wire. Answer in MPa. Answer: Question 6 Not yet answered Marked out of 2.00 Flag question b) The strain in the wire. Answer:
Question 7 Not yet answered Marked out of 2.00 P Flag question c) The increase in the length of the steel wire. Answer in mm. Answer:
d) The ideal effort (P). Answer in N. Answer: Question 5 Not yet answered Marked out of 1.00 Flag question A 40 kg block of aluminum is hung from the end of a vertical 0.50 m long steel wire with a cross-sectional area of 5.0 x 105m? Young's modulus of aluminum: 7.0 x 1010 Pa Young's modulus of steel: 2.0 x 104 Pa Tensile strength of steel: 3.0 x 10* Pa Using the information above, determine the following: a) The stress in the wire. Answer in MPa. Answer: Question 6 Not yet answered Marked out of 2.00 Flag question b) The strain in the wire. Answer:
Question 7 Not yet answered Marked out of 2.00 P Flag question c) The increase in the length of the steel wire. Answer in mm. Answer: