Physics, asked by rashmiisinghrs, 3 months ago

after a fall a 95 kg rock climber finds himself dangling from the end of a rope that had been 15m long and 9.6mm in diameter but has stretched by 2.8dm. for the rope calculate: (i) the strain (ii) the stress.​

Answers

Answered by jolysanthosh57
0

Explanation:

If L(=1500cm) is the unstretched length of the rope and ΔL=28.cm is the amount it stretches, then the strain is :

ΔL/L=(28cm)(1500cm)=1.9×10−3

(b) The stress is given by F/A where F is the stretching force applied to one end of the rope and A is the cross-sectional area of the rope. Here F is the force of gravity on the rock climber. If m is the mass of the rock climber then F=mg. If r is the radius of the rope then A=πr2. Thus the stress is:

AF=πr2mg=π(4.8×103m)2(95kg)(9.8m/s2)=1.3×107N/m2

(c) Young’s modulus is the stress divided by the strain:

$$E = (1.3 \times 10^7N/m^2

) / (1.9 \times 10^{–3}) = 6.9 \times 10^9 N/m^2 $$

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