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
Answer:
if you look it up i think u can find it it could be a mealltiod Co 27
Explanation:
Electrons: negative charge
Protons: positive charge
Neutrons: negative charge
The atom would have to have more electrons than protons
Hope this helps :)
Answer:
A) OA, AB, BC
B) 25m/s^2
C) see explanation
D) 25
E) Rest
Explanation:
From the Velocity time graph shown:
The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.
Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.
-ve slope = BC
B) Acceleration of body in path OA.
Acceleration = change in Velocity / time
Acceleration = (150 - 0) / 6
Acceleration = 150/6 = 25m/s^2
C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).
D) Length of BC
BC corresponds to the distance moved, that velocity / time
Velocity = 150 ; time = 6
Therefore Distance (BC) = 150/6 = 25
E.) Velocity =0 ; Hence body is at rest