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
The temperature will increase
Explanation:
V=40m/s
Vy=V.sina=40.sin20=40 . 0.342=13.68m/s
Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s
Projectile travels during 5 seconds and the ramge becomes:
x=V.t=30.64 . 5=153.2m
If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that
<em>n</em> = 38 N
The friction force is proportional to the normal force by a factor of 0.27, so that
<em>f</em> = 0.27 (38 N) ≈ 10.3 N
and so the answer is D.
