Answer:
The correct answer is A The distance is greater in the first hour because her speed is faster.
Explanation:
During the first hour, Anna is driving at a speed of 50 km/h. During the second hour, she is only driving at a speed of 30 km/h. The faster she goes, the farther she will go.
Hope this helps,
♥<em>A.W.E.</em><u><em>S.W.A.N.</em></u>♥
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
Answer:
Explanation:
Given a school bus.
Let say initially the school bus is traveling with speed "v"
Let assume mass of school bus is "m"
Then, the initial kinetic energy is
K.E_initial = ½mv²
Now, if the initial velocity is tripled,
Then, the new velocity is
v_new = 3v.
Note: the mass of the school does not change it is constant
Then, new kinetic energy is
K.E_new = ½m(v_new)²
v_new = 3v
Then,
K.E_new = ½m(3v)²
K.E_new = ½m × 9v²
K.E_new = 9 × ½mv²
Since K.E = ½mv²
Then,
K.E_new = 9 × K.E
So, the new kinetic energy will be 9 times the initial kinetic energy.
So, option D is correct
D. It will be nine times greater.