According to Hooke's law, a 30-N force stretches a spring by 0.73 m if the spring constant k is such that
30 N = k (0.73 m)
k = (30 N) / (0.73 m)
k ≈ 4.1 N/m
The ball’s speed and direction at the lowest point of trajectory are; 3.1 m/s and swing to the same angle on the other side
<h3>What is the speed at the lowest point?</h3>
We will assume that all of the motion will transfer to Kinetic Energy at the lowest point. Now, the difference in height will be;
Δh = (130 – 130*cos °5) = 0.49 cm
The change in PE is;
ΔP.E = mgΔh
ΔP.E = 9.8m/s² × 0.75kg × 0.049 m
ΔP.E = 0.36015 J
Formula for kinetic energy is;
K.E = ¹/₂*m*v²
From conservation of energy;
K.E = P.E
Thus;
¹/₂* 0.075 * v² = 0.36015
v = √[(0.36015 * 2)/0.075]
v = 3.1 m/s
Thus, the pendulum will swing to the same angle on the other side.
Read more about Speed of Pendulum at; brainly.com/question/17054952
Sorry but didn’t get the question
I am almost positive the answer is B if this helped please give the brainliest
