Answer:
60 rad/s
Explanation:
∑τ = Iα
Fr = Iα
For a solid disc, I = ½ mr².
Fr = ½ mr² α
α = 2F / (mr)
α = 2 (20 N) / (0.25 kg × 0.30 m)
α = 533.33 rad/s²
The arc length is 1 m, so the angle is:
s = rθ
1 m = 0.30 m θ
θ = 3.33 rad
Use constant acceleration equation to find ω.
ω² = ω₀² + 2αΔθ
ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)
ω = 59.6 rad/s
Rounding to one significant figure, the angular velocity is 60 rad/s.
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill,
= 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;

Therefore, the drill's angular displacement during that time interval is 24.17 rad.
Answer:
a) - 72.5°c
b) pressure = 3625.13 Pa
c) density = 0.063 kg/m^3
d) it is a subsonic aircraft
Explanation:
a) Determine Temperature
Temperature at 19.5 km ( 19500 m )
T = -131 + ( 0.003 * altitude in meters )
= -131 + ( 0.003 * 19500 ) = - 72.5°c
b) Determine pressure and density at 19.5 km altitude
Given :
Po (atmospheric pressure at sea level ) = 101kpa
R ( gas constant of air ) = 0.287 KJ/Kgk
T = -72.5°c ≈ 200.5 k
pressure = 3625.13 Pa
hence density = 0.063 kg/m^3
attached below is the remaining part of the solution
C) determine if the aircraft is subsonic or super sonic
Velocity ( v ) =
=
= 283.8 m/s
hence it is a subsonic aircraft
Answer:
u = - 38.85 m/s^-1
Explanation:
given data:
acceleration = 2.10*10^4 m/s^2
time = 1.85*10^{-3} s
final velocity = 0 m/s
from equation of motion we have following relation
v = u +at
0 = u + 2.10*10^4 *1.85*10^{-3}
0 = u + (21 *1.85)
0 = u + 38.85
u = - 38.85 m/s^-1
negative sign indicate that the ball bounce in opposite directon