The elastic potential of a spring can be calculated using the formula
E = (1/2) k x²
where are given that
k = 120 N/m
the stretched length is
x = 0.02 m
The elastic potential is
E = (1/2) (120 N/m) (0.02 m)²
E = 0.024 N-m<span />
I'm going too hell for putting answers that are not really good enough for you but I need my answers and I need help with my answers
Work done = Mass * displacement
W = 50 * 5 = 250 watts
In short, Your Answer would be 250 watts
Hope this helps!
Answer:
The answer to your question is: D) Ф₂ = 49.71°
Explanation:
Data
n₁ = 1.33
Ф₁ = 35°C
n₂ = 1
Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)
Process
Substitution
Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)
Ф₂ = sin⁻¹ (1.33 sin 35/1)
Ф₂ = sin⁻¹ (1.33 x 0.574/ 1)
Ф₂ = sin⁻¹ ( 0.7628 / 1)
Ф₂ = sin⁻¹ (0.7628)
Ф₂ = 49.71°
