Answer:
Therefore maximum stretch is y2 = 32.36 m
Explanation:
In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium
- W = 0
k Δx = mg
k = mg / Δx
k = 80 9.8 / (30-20)
k = 78.4 N / m
now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m
starting point. When will you jump
Em₀ = U = mg y
final point. Just when the rope starts to stretch
= K = ½ m v²
Em₀ = Em_{f}
mg y = ½ m v²
v = √ 2g y
v = √ (2 9.8 20)
v = 19.8 m / s
now all kinetic energy is transformed into elastic energy
starting point
Em₀ = K = ½ m v²
final point
Em_{f} = + U = ½ k y² + m g y
Emo = Em_{f}
½ m v² = ½ k y² + mgy
k y² + 2 m g y - m v² = 0
we substitute the values and solve the quadratic equation
78.4 y² + 2 80 9.8 y - 80 19.8² = 0
78.4 y² + 1568 y - 31363.2 = 0
y² + 20 y - 400 = 0
y = [- 20 ±√ (20 2 +4 400)] / 2
y = [-20 ± 44.72] / 2
the solutions are
y₁ = 12.36 m
y₂ = 32.36 m
These solutions correspond to the maximum stretch and its rebound.
Therefore maximum stretch is y2 = 32.36 m
An object thats NEGATIVELY charged has more electrons than protons, An object thats POSITIVELY charged has less electrons than protons, An object that NOT charged has the same number of electrons and protons.
Matter is radioactive and decays over time
Henri Becquerel discovered that <span>uranium salts spontaneously emit a penetrating radiation that can be registered on a photographic plate.
hope this helps</span>
Answer:
3.55m/s^2
Explanation:
S = ut + 1/2at^2
S = 48m, u = 0, t = 5.2
48 = (0×5.2) + (1/2 × 5.2^2)a
48 = 13.52a
Dividing through by 13.52
a = 48/13.52
a = 3.55m/s^2
Spam email<span> can contain malicious computer code and viruses,it can be an attempt to commit fraud,and it can be an attempt to get personal and financial.</span>