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:
I can't see the picture
Explanation:
Im sorry can you right whats on the picture down in the comments plz.
Answer:
34.6 m/s
Explanation:
From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg
Final mass will be 31.5+25.9=57.4 kg
From formula of momentum
M1v1=m2v2
Making v2 the subject of the formula then
Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then
Answer: here you go I was looking for this answer everywhere,I have it now so it’s 6.30 x 10^-7 s
Explanation:
I hope this helps☺️
<span>Light can travel in a vacuum, and ... strange as it may seem ...
its speed is always the same, even if the light source is moving. </span>
