Answer:
4.8 turns would be made around the tube
Explanation:
You need the circumference of the tube since it's just a lifted circle
2×pi×3=18.85
90.42/18.85=4.79
And round that to 4.8
Answer:
Answer:
4 ms
Explanation:
initial velocity, u = 75 m/s
final velocity, v = 0
distance, s = 15 cm = 0.15 m
Let the acceleration is a and the time taken is t.
Use third equation of motion
v² = u² + 2 a s
0 = 75 x 75 - 2 a x 0.15
a = - 18750 m/s^2
Use first equation of motion
v = u + at
0 = 75 - 18750 x t
t = 4 x 10^-3 s
t = 4 ms
thus, the time taken is 4 ms.
Explanation:
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
All planets move around the sun
Answer:
r = 2161.9 m
Explanation:
Aerodynamic lift(L) is perpendicular to the wing, which is tilted 40 degrees to the horizontal.
Since the plane is moving in a horizontal circle, the vertical component of the lift must cancel the weight W of the airplane, but the horizontal component is the centripetal force that keeps it in a circle.
L is perpendicular to wing at angle θ with respect to horizontal
Thus,
Vertical component of lift is:
L cosθ = W = mg
Thus, m = L cosθ / g - - - - (eq1)
Horizontal component of lift is:
L sinθ = centripetal force = mv² / r - - - - (eq2)
Combining equations 1 and 2,we have;
L sinθ = (L cosθ / g)(v² / r)
L cancels out on both sides to give;
tanθ = v²/ rg
r = v² / (g tanθ)
We are given;
velocity; v = 480 km/hr = 480 x 10/36 = 133.33 m/s
r = 133.33²/[(9.8) tan(40)] = 2161.9 m
