3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming t
1 answer:
Answer:
Work done is 12.3 J
Explanation:
We have,
Mass of puck, m = 0.35 kg
Force of friction acting on the puck when it slides is 0.15 N
Distance travelled by the puck is 82 m.
It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by
The work done on the puck is 12.3 J.
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Answer:
C
Sign-Negative
Explanation:
We are given that
Electric field = (Radially downward)
Acceleration=(Upward)
Mass of charge=3 g=kg
1kg=1000g
We have to find the magnitude and sign of charge would have to be placed on a penny .
By newton's second law
Substitute the values then we get
Substitute the values then we get
C
Sign of charge =Negative
Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.