at the end of a 1.9 m string swings around in a circle. if the object\'s centrip

Question

at the end of a 1.9 m string swings around in a circle. if the object's centripetal acceleration is 18 m/s, then what is its linear speed? a. 1.458 m/s b. 1.849 m/s d. 1.542 m/s 1.556 m/s 1.805 m/s 1.774 m/s h object moving in a circle has speed 21,m/s and centripetal acceleration 3.9,m/s. Find the radius of the circle. 12. a. 140.9.m b. 90.95 m 110.3,m d. 113.1.m e 90.42,m f. 73.03.m A 2.2 m long rod has mass 8.0 kg. If the rotation axis is located at one end of the rod, then find its moment of inertia. 13. a. 9.805 kg m b. 13.22 kg-m* c. 12.44 kg-m d. 15.37 kgr e. 14.84 kg-m2 f. 12.91 kg m2 A 300-kg disk (with uniform density) is shown below. Its rotation axis is through its center. If the disk is initially at rest, how long does it take to reach an angular velocity of 15 rad/s? 420 N 14. 18 N a. 72.58s b. 88.57s . 64.95s d. 78.04s e. 85.69 s f. 87.94 s 0.75 m i 1.0 m 16N 010 Spinning Disk 01 (SN A 2000.kg solid disc with radius 0.36,m is at rest when a 170.m-N torque is applied to the wheel. How long will it take the disc to reach an angular speed of 5600,rev/min? 15. a. 14.5 min b. 7.663 min c. 13.35,min d. 7.451,min e. 13.66 min f. 7.05.min

Explanation / Answer

11.

Centripetal acceleration

ac=v2/r

Linear speed is

v=sqrt(a*r)=sqrt(1.9*1.8)

v=1.849 m/s

12.

radius

r=v2/a=212/3.9

r=113.1 m

13.

Moment of inertia of rod fixed at one end is

I=(1/3)ML2=(1/3)*8*2.22

I=12.91 kg-m2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.