Position: x = 18t y = 4t - 4.9t²
First derivative: x' = 18 y' = 4 - 9.8t
Second derivative: x'' = 0 y'' = - 9.8
Position vector: P = (18t) i + (4t - 4.9t²) j
Velocity vector: V = (18) i + (4 - 9.8t) j
Acceleration vector A = (- 9.8) j
Answer:
F₂= 210 pounds
Explanation:
Conceptual analysis
Hooke's law
Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:
F= K*x Formula (1)
Where;
F is the magnitude of the force applied to the spring in Newtons (Pounds)
K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)
x the elongation of the spring (inch)
Data
The data given is incorrect because if we apply them the answer would be illogical.
The correct data are as follows:
F₁ =80 pounds
x₁= 8 inches
x₂= 21 inches
Problem development
We replace data in formula 1 to calculate K :
F₁= K*x₁
K=( F₁) / (x₁)
K=( 80) / (8) = 10 pounds/ inche
We apply The formula 1 to calculate F₂
F₂= K*x₂
F₂= (10)*(21)
F₂= 210 pounds
The period of the wave would be halved
Answer: D
Explanation: there is less light at that point.
