Find the gravitational force of attraction between a 60-kg woman and a 73-kg man
ID: 1475352 • Letter: F
Question
Find the gravitational force of attraction between a 60-kg woman and a 73-kg man when they are 1.1 m apart (answer in N)
Two children sit 3.20 m apart on a very low-mass horizontal seesaw with a movable fulcrum. The child on the left has a mass of 29.0 kg, and the child on the right has a mass of 38.0 kg. At what distance, as measured from the child on the left, must the fulcrum be placed in order for them to balance? (answer in m)
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Two children sit 3.20 m apart on a very low-mass horizontal seesaw with a movable fulcrum. The child on the left has a mass of 29.0 kg, and the child on the right has a mass of 38.0 kg. At what distance, as measured from the child on the left, must the fulcrum be placed in order for them to balance? (answer in m)
An engine’s flywheel can be modeled as a uniform solid cylinder of mass 5.6 kg and radius 0.19 m. Find its angular velocity when its angular momentum is 8.7 kg·m2/s. (answer in rad/s)Explanation / Answer
m1 = 60 kg , m2 =73 kg , d =1.1 m
G=6.6x10^-11 N.m^2/kg^2
From Newton gravitaional law F =Gm1m2/d^2
F = (6.67x10-11x60x73)/(1.1x1.1)
F =2.414x107 N
m1 =29 kg , m2 = 38 kg, L =3.2 m
Net torque about an axis passing through fulcrum is zero.
r1xF1 =r2XF2
dm1g = (L-d) m2g
dx29 = (3.2-d)38
d=1.815 m
The fulcrum is placed 1.815 m from left end
m =5.6kg , r =0.19m
L =8.7kg.m^2/s
Moment of inertia of wheel I =mr^2
Angular momentum L = Iw
w =L/I = L/mr^2 = (8.7)/(5.6x0.19x0.19)
Angular velocity w = 43.04 rad/s
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