B49 Explain the terms tensile stress and teale strain as applied Lo a specimen of material and explain the meaning of the word tensile. A car breaks down and the driver asks a lieal to tow it using a piece of nylon rope whils 10.00 m long and has a diameter of 10.m The rope obeys Hooke's law and has a Ynmg modulus of 3.0 x 10°Nm. The mass of the car and driver is 750 kg. (a) When towing on a level road at a cunntant speed, it is found chat the rope cstenth by 0.025 m. Calculate the tension in ilepe and hence the net resistive force aclug the towed cur. (b) The two cars now ascend a slope whul rises 1.0m vertically for every 1'5m travelled on the road. They maintan the same speed as in part (a). What is the new length of the towrope? (c) How much clastic energy is stored in ihe rope while the cars are climbing? (d) A stretched towrope must be regarded aN

Q: The material for the 58-mm-long specimen has the stress–strain diagram shown. A) If P = 150 kN is…

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Q: An extremely expensive piece of spider silk is 2.5 m long and 0.5 mm in diameter is anchored to the…

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Q: Q2.9/ A copper wire with a circular cross section of a diameter (d31.2 mm) was cast by a force of…

A: From the question, it is given that- Force on copper wire: F = 90 N Diameter of copper wire: d = 1.2…

Q: A car breaks down and the driver asks a lal to tow it using a piece of nylon rope wha. ls 10.00 m…

A:

Q: A disk between vertebrae in the spine is subjected to a shearing force of 570 N. Find its shear…

A: The expression for the shear deformation, S=FL0∆xπd24∆x=4FL0πSd2

Q: A structural steel rod has a radius of 10 mm and a length of 1.0 m. A100 kN force stretches it along…

A: Calculating the stress in the rod:

Q: Many caterpillars construct cocoons from silk, one of the strongest naturally occurring materials…

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Q: A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0…

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Q: Which one of the following cannot be a definition of an elastic modulus? (A) AP -V; AV Fh B AAx FLi…

A: In the given question, We have to choose the correct option, i.e, which can not be a defination of…

Q: After a fall, a 110 kg rock climber finds himself dangling from the end of a rope that had been 19 m…

A:

Q: A 29.0 m long aluminum cable with a cross-sectional area of 2.01 x 10- m is used to suspend a 3.20 x…

A: The expression for young’s modulus of the material is, \Y=FAΔLL0ΔL=FL0AY=mgL0AY

Q: 1. In her gymnastic routine, Regina dismounts from the uneven-parallel bars and lands…

A: stress modulus E=FLA△L F=EA△LL

Q: A 2.7-m-tall, 50-cm-diameter concrete column supports a 4.0×105 kg load. By how much is the column…

A: Given:Length of the column, L = 2.7 mDiameter of the column, D = 50 cmRadius of the column, R = 25…

Q: One end of a vertical metallic wire of length 2.0 m and diameter 1.0 mm is attached to a ceiling,…

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Q: A 2.00-m-long, 1.00-mm-diameter wire is suspended from the ceiling. Hanging a 4.50 kg mass from the…

A: According to question--- We have to find--- length = 2m. diameter = 1mm. mass = 4.5 kg. g = 10 m/…

Q: Humans can bite with a force of approximately 810 N. If a human tooth has the Young's modulus of…

A: SOlution: as given F = 810 N E = youngs modulus of bone= A = 1.5 cm2 = 1.5x10-4 m2 L = 1.5 cm =…

Q: One end of a metal rod of radius R =8.7 mm and length L-81 cm is held in a vise, see figure. A force…

A: The expression for the Young’s modulus is,

Q: A 90 kg mountain climber stretches her 1.0 cm diameter nylon rope by 1.3 m when she hangs below a…

A: Find the force exerted on the rope. F=mg=90 kg9.8 m/s2=882 N Find the cross sectional area of the…

Q: In normal motion, the load exerted on the hip joint is 2.5 times body weight. (a) Calculate the…

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Q: After a fall, a 73 kg rock climber fınds himself dangling from the end of a rope that had been 17 m…

A: Given  Original length L=17m Change in length ∆L=3.3 cm =0.033m Mass m=73 kg  Dia d=7.5 mm

Q: A wall climber of mass 85 kg dangles at the end of a rope. The rope has a diameter of 9.6 mm, and is…

A: mass= 85 Kg Length, L= 21 m radius, r= 4.8 mm ∆L= 3.2 cm

Q: Assume Young's modulus for bone is 1.50 x 1010 N/m2. The bone breaks if stress greater than o= 79104…

A: maximum stress= sigma= 79104 N/m2

Q: A 10.2 m long wire with a cross-sectional area of 0.126 cm2 stretches by 0.196 cm when a tensile…

A: The given values are, L=10.2 mA=0.126 cm2=0.126×10-4 m2∆L=0.196 cm=0.196×10-2 mF=6250 N

Q: A man with a mass of 72.1 kg stands on one foot. His femur has a cross-sectional area of 8.00 cm2…

A: Given, M=72.1 kgg=9.81 m/s2W=Mg=72.1×9.81=707.301 NA=8.00 cm2=8×10-4 m2L=49.6 cm=0.496 mE=9.40×109…

Q: 1. Assume Young’s modulus for bone is 1.50 x 1010 N/m2. The bone breaks if stress greater than 1.50…

A: The maximum force is, F=stressArea=stressπd24

Q: A wire has 20 x 10^10 N/msquared young modulus. Its diameter is 1mm that can handle a tension of 0.2…

A: Given data- young modulus= 20×1010 N/m2 1mm diameter can handle a tension of 0.2 kN.

Q: Assume Young's modulus of a femure bone of an effective diameter 2.5 cm is 1.5 x1010 N/m2. The bone…

A: Given: A femur bone of an effective diameter 2.5 cm and it's Young's modulus = 1.5 × 1010 N/m2.…

Q: A 15.9 cm long tendon is found to stretch 3.81 mm when a force of 18.0 N is applied to it. If the…

A: The length of the tendon is l = 15.9 cm. The force applied on the tendon is F = 18.0 N. The average…

Q: A 1.3 x 10-3-kg house spider is hanging vertically by a thread that has a Young's modulus of 4.4 x…

A:

Q: Show transcribed image text For safety in climbing, a mountaineer uses a nylon rope that is 51m long…

A: Length, L = 51 m  elongation, l = 2.1 m mass, m = 110 kg diameter, d = 1.1 cm

Q: Assume Young's modulus for bone is 1.50 x 1010 N/m. The bone breaks if stress greater than 0= 80179…

A: Sigma= 80179 N/m2

Q: After a fall, a 94 kg rock climber finds himself dangling from the end of arope that had been 19 m…

A: Given value--- mass = 93 kg. initial length = 19 m. initial diameter = 8.4 mm. Change in length =…

Q: A 28.0 m long aluminum cable with a cross-sectional area of 2.01 ✕ 10−3 m2 is used to suspend a 3.40…

A: Y=mg/A∆L/L   =mgLA∆L∆L=mgLYA

Q: Find the minimum diameter of a (50 m)- long nylon string that will stretch no more than (1 cm) when…

A: Initial length of the nylon string Lo = 50 m  Change in length of the nylon string ∆L = 1 cm =>…

Q: After a fall, a 80 kg rock climber finds himself dangling from the end of a rope that had been 14m…

A:

Q: After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m…

A: Part (a) extension or change in length= 3.3cm = 0.033 m  length = 16m  Strain = change in length ÷…

Q: A vertical steel girder with a crosssectional area of 0.15 m2 has a 1550 kg sign hanging from its…

A: Given: Area of cross section A of the steel girder is 0.15 m2. Mass M hanging at the end is of 1550…

Q: After a fall, a 88 kg rock climber finds himself dangling from the end of a rope that had been 12 m…

A:

Q: A bone in a woman's leg has an effective cross-sectional area of 3.07 cm². If the bone is 36.1 cm…

A: Compressive strength of human bone is CS= Cross- sectional area of the bone, A= 3.07 cm2 = Length of…

Q: After a fall, a 94 kg rock climber finds himself dangling from the end of a rope that had been 16 m…

A:

Q: A car breaks down and the driver asks a likal to tow it using a piece of nylon rope whin l 10.00 m…

A:

Q: Stress on a Mountaineer’s Rope. A nylon rope used by mountaineers elongates 1.10 m under the weight…

A: Given: The elongation is 1.10 m. The mass of the climber is 65.0 kg. The length of the rope is 45.0…

Q: " The femur is a bone in the leg whose minimum cross-sectional area is about 4.0 X 10-ª m². A…

A:

Q: Assume Young’s modulus for bone is 1.40×1010 N/m2 . The bone breaks if stress greater than 1.50 ×…

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Q: After a fall, a 100 kg rock climber finds himself dangling from the end of a rope that had been 14 m…

A: GivenMass of rock climber, m=100 kgOrignal Length of rope, L=14 mDiameter of rope, D=10 mm=10×10-3…

Q: A 16.3 cm long tendon is found to stretch 2.11 mm when a force of 18.8 N is applied to it. If the…

A: Given value--- initial length = 16.3 cm. change in length = 2.11 mm. Force = 18.8 N . diameter =…

Q: What hanging mass will stretch a 2.2-m-long, 0.34 mm - diameter steel wire by 1.2 mm? The Young's…

A:

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