Centrii Heeellp.. (1 Viewer)

shantu1992

Member
Joined
Apr 23, 2008
Messages
154
Gender
Male
HSC
2009
1. A mass of 1.25 kg moves in a circular path with a constant speed of 3.15 ms-1 on a horizontal frictionless surface. the mass is constrained to move on the circular path by a light string of length 2.35 m that has one end fixed and the other end attatched to the mass. calculate the tension in the string..

2. an ice skater moves in a circular path pf radiues 6.8 m with a speed of 6 ms-1. what centripetal acceleration does she experience

3. calculate the speed of a satellite in an orbit around the earth at a distance of 7.5 * 10 to the power of 6 meters from the earths centre. the mass of teh earth is 5.98 * 10 tot he power of 24 kg

thanks heaps
 

Kings407

Member
Joined
Jan 29, 2008
Messages
37
Gender
Male
HSC
2008
shantu1992 said:
1. A mass of 1.25 kg moves in a circular path with a constant speed of 3.15 ms-1 on a horizontal frictionless surface. the mass is constrained to move on the circular path by a light string of length 2.35 m that has one end fixed and the other end attatched to the mass. calculate the tension in the string..

2. an ice skater moves in a circular path pf radiues 6.8 m with a speed of 6 ms-1. what centripetal acceleration does she experience

3. calculate the speed of a satellite in an orbit around the earth at a distance of 7.5 * 10 to the power of 6 meters from the earths centre. the mass of teh earth is 5.98 * 10 tot he power of 24 kg

thanks heaps
1)
Tension in the string = centripetal force
Fc=(mv^2)/r m=1.25kg
v=3.15m/s
r=2.35m
F=(1.25*3.15^2)/2.35
=5.2779....N
=5.28N

2)
F=ma
a=F/m
=(v^2)/r
=(6^2)/6.8
=5.294...m/s/s
=5.29m/s/s
3)
v=square root (GMe/d)
=square root (6.67x10^-11x5.98x10^24/7.5x10^6)
=7292.60813...m/s
=7293m/s



Hope this helps altho 3) mite be wrong coz its so large
 

BDinu1

New Member
Joined
Nov 13, 2008
Messages
1
Gender
Female
HSC
2011
nope i checked it, 3 is right. You guys are really smart :D im still in highschool grade 9 so im not as advance... :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top