here,
the length of rod, l = 4 m
cross sectional area , A = 0.05 m^2
dl = 0.01 m
tension , T = 5000 N
the Young's modelus , Y = stress /strain
Y = ( T /(A))/(dl/l)
Y = (5000 * 4)/(0.05 * 0.01)
Y = 4 * 10^7 N/m^2
the Young's modulus is 4 * 10^7 N/m^2
A metal rod that is 4.0 m long and 0.05 m2 in cross-sectional area is found to stretch 0.01 m under a tension of 50...
Exercise 11.27 A metal rod that is 5.00 m long and 0.49 cm2 in cross-sectional area is found to stretch 0.13 cm under a tension of 4300 N . Part A What is Young's modulus for this metal? Express your answer using two significant figures. Y = Pa SubmitMy AnswersGive Up
Exercise 11.27 A metal rod that is 5.00 m long and 0.59 cm2 in cross-sectional area is found to stretch 0.25 cm under a tension of 4300 N . Part A What is Young's modulus for this metal? Express your answer using two significant figures. Y = Pa SubmitMy AnswersGive Up Incorrect; Try Again; 3 attempts remaining
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Figure < 1 of 1 Consider, for instance, a bar of initial length L and cross-sectional area A stressed by a force of magnitude F. As a result, the bar stretches by AL (Figure 1) Let us define two new terms: • Tensile stress is the ratio of the stretching force to the cross-sectional area: stress = 5 • Tensile strain is the ratio of the elongation of the rod to the initial length of the bar strain= 41 It...
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