Asked by tony
A short 10-g string is used to pull a 50-g toy across a frictionless horizontal surface. If a
3.0 x 10^(–2)-N force is applied horizontally to the free end, the force of the string on the toy, at the other end, is:
A) 0.15 N
B) 6.0 x 10^(–3) N
C) 2.5 x 10^(–2) N
D) 3.0 x 10^(–2) N
E) 3.5 x 10^(–2) N
3.0 x 10^(–2)-N force is applied horizontally to the free end, the force of the string on the toy, at the other end, is:
A) 0.15 N
B) 6.0 x 10^(–3) N
C) 2.5 x 10^(–2) N
D) 3.0 x 10^(–2) N
E) 3.5 x 10^(–2) N
Answers
Answered by
bobpursley
acceleration=force/total mass=.03N/.06kg= .5m/s^2
forceonthetoy=mass*a=.05*.5=.025N
forceonthetoy=mass*a=.05*.5=.025N
Answered by
j
bob is right, the answer is C
Answered by
nick acock
a is constant on both thus finding a will answer the question
acceleration=force/total mass=.03N/.06kg= .5m/s^2
forceonthetoy=mass*a=.05*.5=.025N
acceleration=force/total mass=.03N/.06kg= .5m/s^2
forceonthetoy=mass*a=.05*.5=.025N
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