# 1.A vault is opened by applying a force of 300 N perpendicular to the plane of the door, 0.80 m from the hinges. Find the torque due to this force about an axis through the hinges.a. 120 N⋅mb. 240 N⋅mc. 300 N⋅m d. 360 N⋅m2.A meter stick is supported by a knife-edge at the 50-cm mark. Doug hangs masses of 0.40 and 0.60 kg from the 20-cm and 80-cm marks, respectively. Where should Doug hang a third mass of 0.30 kg to keep the stick balanced?a. 20cmb. 70cm c. 30cm d. 25cm3.A copper wire of length 2.0 m, cross sectional area 7.1 × 10−6 m2 and Young's modulus 11 × 1010 N/m2 has a 200-kg load hung on it. What is its increase in length? (g = 9.8 m/s2)a. 0.50 mmb. 1.0 mmc. 2.5 mm d. 5.0 mm

Question
Asked Dec 12, 2019
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1.A vault is opened by applying a force of 300 N perpendicular to the plane of the door, 0.80 m from the hinges. Find the torque due to this force about an axis through the hinges.
a. 120 N⋅m
b. 240 N⋅m
c. 300 N⋅m d. 360 N⋅m
2.A meter stick is supported by a knife-edge at the 50-cm mark. Doug hangs masses of 0.40 and 0.60 kg from the 20-cm and 80-cm marks, respectively. Where should Doug hang a third mass of 0.30 kg to keep the stick balanced?
a. 20cm
b. 70cm c. 30cm d. 25cm
3.A copper wire of length 2.0 m, cross sectional area 7.1 × 10−6 m2 and Young's modulus 11 × 1010 N/m2 has a 200-kg load hung on it. What is its increase in length? (g = 9.8 m/s2)
a. 0.50 mm
b. 1.0 mm
c. 2.5 mm d. 5.0 mm

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## Expert Answer

Step 1

1.Write the expression for torque.

Step 2
1. The third mass is 0.30kg must be on the left side. The torque about the pivot is equal to zero. Let x be the distance from left end.
Step 3

Thus option (c) is the correct answer. Thus Dong shou...

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