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:
0.02 s
Explanation:
Take the (+x) direction to be up.
The average velocity v during a time interval Δt is the displacement Δx divided by Δt.
v=Δx/Δt
=x_f-x_i/t_f-t_i (1)
We assume that your height is 1.6m
Solving [1]
Δt=Δx/v
= 0.02 s
I only know P and V and P is pressure and V is volume
Answer:
Explanation:
Given that,
The volume of the balloon is
V = 440 × 10³ m³
Buoyant force F?
Given the density of the surrounding to be 2.58 kg/m³
ρ = 2.58 kg/m³
The buoyant force is the weight of water displaced and it is calculated using
F_b = ρVg
Where
F_b is buoyant force
ρ is density
V is the volume of the liquid displace.
g is the acceleration due to gravity
Then,
F_b = ρVg
F_b = 2.58 × 440 × 10³ × 9.81
F_b = 1.1 × 10^7 N
Answer:
Speed of the this part is given as

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate
Explanation:
As we know by the momentum conservation of the system
we will have

here we know that

the momentum of two parts are equal in magnitude but perpendicular to each other
so we will have


now from above equation we have



Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate