Answer:
0.045 J
Explanation:
From the question,
The elastic potential energy stored in a spring is given as,
E = 1/2ke²...................... Equation 1
Where E = elastic potential energy, k = spring constant, e = compression.
Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m
Substitute these values into equation 1
E = 1/2(100)(0.03²)
E = 50(9×10⁻⁴)
E = 0.045 J
Hence the right option is 0.045 J
^^^^^^^^^^^^^^^^^^^^^^^^^^^ is correct
False... I hope that helps ;)
Answer: Here this will help you..
Explanation:
1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second
5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second
10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second
20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second
30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second
40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second
50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second
75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second
100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second
