Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
That is correct. Or so I believe. Either more or less than the other on the amount of protons and electrons, you can get either an unstable or a stable atom of an element.
Given: Universal Gravitational constant = G = 6.67 x 10⁻¹¹ N m²/Kg²
Mass₁ = 70 Kg; Mass₂ = 70 Kg Radius r = 1.5 m; Force F = ?
Formula: F = Gm₁m₂/r²
F = (6.67 x 10⁻¹¹ N m²/Kg²)(70 Kg)(70 Kg)/(1.5 m)²
F = 1.45 X 10⁻⁷ N