Answer:
θ = 6.3 *10³ revolutions
Explanation:
Angular acceleration of the drill
We apply the equations of circular motion uniformly accelerated
ωf= ω₀ + α*t Formula (1)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (s)
Data
ω₀ = 0
ωf = 350000 rpm = 350000 rev/min
1 rev = 2π rad
1 min= 60 s
ωf = 350000 rev/min =350000*(2π rad/60 s)
ωf = 36651.9 rad/s
t = 2.2 s
We replace data in the formula (2) :
ωf= ω₀ + α*t
36651.9 = 0 + α* (2.2)
α = 36651.9 / (2.2)
α = 17000 rad/s²
Revolutions made by the drill
We apply the equations of circular motion uniformly accelerated
ωf²= ω₀ ²+ 2α*θ Formula (2)
Where:
θ : Angle that the body has rotated in a given time interval (rad)
We replace data in the formula (2):
(ωf)²= ω₀²+ 2α*θ
(36651.9)²= (0)²+ 2( 17000 )*θ
θ = (36651.9)²/ (34000 )
θ = 39510.64 rad = 39510.64 rad* (1 rev/2πrad)
θ = 6288.31 revolutions
θ = 6.3 *10³ revolutions
Answer:
When the temperature decreases the particals start to slow down.
True
The half-life isn’t applicable to a first order reaction because it does not rely on the concentration of reactant present. However the 2nd order reaction is dependent on the concentration of the reactant present.
The relationship between the half life and the reactant is an inverse one.
The half life is usually reduced or shortened with an increase in the concentration and vice versa.
