Question

a stone is tied to one end of a string and whirled in a horizontal circle of a radius 1 m at 20 revolution per minute calculate the angular velocity and linear speed of the stone also find the centripetal acceleration​

Answer 1

• Answer:
• Answer:Given length of the string l= 1 m also angular velocity ω=
• Answer:Given length of the string l= 1 m also angular velocity ω= 44
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×r
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v=
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 44
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/s
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c =
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = r
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv 2
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv 2
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv 2
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv 2
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv 2 Thus a
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv 2 Thus a c
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv 2 Thus a c
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv 2 Thus a c =0.25m/s
• Answer:Given length of the string l= 1 m also angular velocity ω= 4422 revolution/second.We know linear velocity v=ω×rThus velocity v= 4422 ×1=0.5 m/sAnd we know centripetal or radial acceleration a c = rv 2 Thus a c =0.25m/s 2