I think the question should be the below:
<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:
<span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>
<span>x (max) = a(max) /ω² </span>
<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>
Since the direction of the force and the direction of the path is perpendicular, the person is not doing any physical work.
P = IV
I = P/V = 30 / 120 = 0.25 A.
Current = 0.25A
Answer:
Option B
Explanation:
<h3>According to Newton's third law, for every reaction there will be equal and opposite reaction</h3>
Here in this case the force of the club hitting the golf ball will be in one direction and the force acting on club due to golf ball will be in opposite direction and magnitude of this force will be same as the magnitude of the force of the club hitting the golf ball
In this case the action will be the force of the club hitting the golf ball and reaction will be the force acting on club due to golf ball
∴ The club pushes against to golf ball with a force equal and opposite to the force of the golf ball on the club
