Answer:
The angular displacement of the blade is 576,871.2 radians
Explanation:
Given;
angular speed of the Helicopters rotor blades, ω = 510 rpm (revolution per minute)
time of motion, t = 3 hours
The angular speed of the Helicopters rotor blades in radian per second is given as;
The angular displacement in radian is given as;
θ = ωt
where;
t is time in seconds
θ = (53.414)(3 x 60 x 60)\\
θ = 576,871.2 radians
Therefore, the angular displacement of the blade is 576,871.2 radians
Answer:
SRY I CAN'T ANS :( and hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
I believe that B is the answer.
Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mass hangs vertically from it?
