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:
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Given:
The mass of the halfback is m = 107 kg
The speed of the halfback is v = 8 m/s
To find the momentum.
Explanation:
The momentum of the halfback is

Thus, the momentum of the halfback is 856 kg m/s
Answer:
translation
Explanation:
If a ball has an approximate mass of 0.5 kg, with what force must be kicked to give it an acceleration of 1.5m / s2?
Third model shows how a comet's tail changes during its orbit...
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