a) r = d / 2 = 1.4 / 2 = 0.7 cm
force = stress * area
= 4.00 * 108 * pi * (0.7 * 10-2)2= 6.157 * 104 N
b) surface area = pi * 0.015 * 0.580 * 10-2 = 2.73 * 10-4
force = stress * surface area
= 4.00 * 108 * 2.73 * 10-4= 1.09 * 105 N
Need Help? Raxd it 3. O 0/2 points I s Answers SerCP10 9.P.012 MI FB Assume...
10. (S&J 12-21) Assume if the shear stress in steel exceeds about 4.00 x 10 N/m, the steel ruptures. Determine the shearing force necessary to: a. shear a steel bolt 1.00 cm in diameter; and b. punch a 1.00-cm-diameter hole in a steel plate 0.500 cm thick.
10. (S&J 12-21) Assume if the shear stress in steel exceeds about 4.00 x 10 N/m, the steel ruptures. Determine the shearing force necessary to: a. shear a steel bolt 1.00 cm in...
Assume if the shear stress in steel exceeds about 4.00 108 N/m2 the steel ruptures. (a) Determine the shearing force necessary to shear a steel bolt 0.75 cm in diameter. N (b) Determine the shearing force necessary to punch a 2.60-cm-diameter hole in a steel plate 0.550 cm thick.
Assume if the shear stress in steel exceeds about 4.00 108 N/m2 the steel ruptures. (a) Determine the shearing force necessary to shear a steel bolt 1.25 cm in diameter. (b) Determine the shearing force necessary to punch a 2.60-cm-diameter hole in a steel plate 0.590 cm thick.
Assume that if the shear stress in steel exceeds about 4.0o x 108 N/m2, the steel ruptures. (a) Determine the shearing force necessary to shear a steel bolt 1.45 cm in diameter. (b) Determine the shearing force necessary to punch a 1.05-cm-diameter hole in a steel plate 0.800 cm thick. Step 1 (a) we know that force F is equal to the surface area A times stress σ so we have 4 108 N/m 2 Your response differs from the...
5. Certain stainless steel has a shear stress limit 4.00x108 N/m2. Determine the shearing forces needed to: (a) Shear a stainless steel bolt 1.00cm in diameter (hint: force parallel to the top&bottom of the bolt) 10000 7 = 3.14.104N (b) Punch a 1.00 cm diameter hole in a stainless steel plate 0.50cm thick (hint: force parallel to the side of the hole)
O 444/6.66 points Previous Answers SerCP10 9.P016 A high-speed lifting mechanism supports a(n) 820-kg object with a steel cable that is 40.0 m long and 4.00 cm in cross-sectional area. (a) Determine the elongation of the cable. (Enter your answer to at least two decimal places.) 4.018 mm (b) By what additional amount does the cable increase in length if the object is accelerated upwards at a rate of 3.5 m/s ? X Enter a number. differs significantly from the...
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14.- From the next exercise obtained the solution From the next e A component is attached to a wall with four 0,375-in.- diameter bolts (each acting in single shear). A load P is applied to the component as shown. The ultimate shearing stress for each bolt is 40 ksi. If the factor of safety is 2, determine the largest load P that can be applied to the...
3 problems, physics
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O-8.66 points SerCP10 2 P025 O Ask Your Teacher My Notes A steam catapult launches a jet aircraft from the aircraft carrier John C. Stennis, ghving it a speed of 155 mi/h in 2.40 s (a) Find the average acceleration of the plane. m/s2 (b) Assuming the acceleration is constant, find the distance the plane moves Need Help? Read it O-6.00 points SerCP10 2 P.027MLFB My Notes O Ask...
+3 points SerCP10 13.P.037 A pendulum clock that works perfectly on Earth is taken to the Moon. Assume that the free-fall acceleration on the Moon is 1.63 m/s (a) Does it run fast or slow there? O fast O slow Neither, it runs the same as on Earth. (b) If the clock is started at 12:00 midnight, what will it read after 15.5 h (Enter the time to the nearest minute.) AM +-4 points SerCP11 13.5.P.039.MI The free-fall acceleration on...
10. + 0/3 points | Previous Answers SerPSET9 12.P.021. My Notes + Ask Your Teacher John is pushing his daughter Rachel in a wheelbarrow when it is stopped by a brick 8.00 cm high (see the figure below). The handles make an angle of a = 20.0° with the ground. Due to the weight of Rachel and the wheelbarrow, a downward force of 405 N is exerted at the center of the wheel, which has a radius of 15.0 cm....