C is true, and just one of those has as much mass as about 1,840 electrons.
Answer:
Calculate the work done by a 47 N force pushing a 0.025 kg pencil 0.25 m ... A boy on a bicycle drags a wagon full of newspapers at 0.80 m/s for 30 min ... A power mower does 9.00 x 105 J of work in 0.500 h. ... p: W 2200ch: w will320,000 T/ ... How much electrical energy (in kilowatt hours) would a 60.0 W light bulb use in ..
Explanation:
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