That it will erupt upon contact. Hope it helps!
Answer:
D.
Explanation:
8.1X10 to the power of 6 Joules.
Answer:
μ = 0.725
Explanation:
This problem refers to Newton's second law.
F = ma
Let's write the equations on each axis
Y Axis
N-W = 0
N = W
N = mg
X axis
F-fr = ma
With the body not started moving its acceleration is zero
F-fr = 0
F = fr
The friction force equation is
fr = μ N
fr = μ m g
Let's replace and calculate
F = μ m g
μ = F / mg
μ = 321 /45.2 9.8
μ = 0.725
Answer:
The answer to the question is
The ladybug begins to slide
Explanation:
To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same
Where the frictional force equals
= μ×N = m×g×μ
and the centripetal force is given by m·ω²·r
If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have
m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁
and for the gentleman bug we have
m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂
But r₁ = 2×r₂
Therefore substituting the values of r₁ =2×r₂ we have
g×μ = ω²·r₁ = g×μ = ω²·2·r₂
Therefore ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide
The ladybug begins to slide