Answer:
The angular acceleration α = 14.7 rad/s²
Explanation:
The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod
Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.
So Iα = Wr
Substituting the value of the variables, we have
mL²α/12 = mgL/2
Simplifying by dividing through by mL, we have
mL²α/12mL = mgL/2mL
Lα/12 = g/2
multiplying both sides by 12, we have
Lα/12 × 12 = g/2 × 12
αL = 6g
α = 6g/L
α = 6 × 9.8 m/s² ÷ 4.00 m
α = 58.8 m/s² ÷ 4.00 m
α = 14.7 rad/s²
So, the angular acceleration α = 14.7 rad/s²
The answer is B. magnesium I am pretty sure
Answer:
a) ![(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]](https://tex.z-dn.net/?f=%28Qa%2Ag%2AVb%29-%28Qh%2AVb%2Ag%29%3D%28Qh%2AVb%2Aa%29%5C%5Cwhere%20%5C%5Cg%3Dgravity%20%5Bm%2Fs%5E2%5D%5C%5Ca%3Dacceleration%20%5Bm%2Fs%5E2%5D)
b) a = 19.61[m/s^2]
Explanation:
The total mass of the balloon is:
![massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\](https://tex.z-dn.net/?f=massball%3Ddensityheli%2Avolumeheli%5C%5C%5C%5Cmassball%3D0.41%20%5Bkg%2Fm%5E3%5D%2A0.048%5Bm%5E3%5D%5C%5Cmassball%3D0.01968%5Bkg%5D%5C%5C%5C%5C)
The buoyancy force acting on the balloon is:
![Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]](https://tex.z-dn.net/?f=Fb%3Ddensityair%2Agravity%2Avolumeball%5C%5CFb%3D1.23%5Bkg%2Fm%5E3%5D%2A9.81%5Bm%2Fs%5E2%5D%2A0.048%5Bm%5E3%5D%5C%5CFb%3D0.579%5BN%5D)
Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.
In the attached image we can see the free body diagram and the equation deducted by Newton's second law
Hope this helps!!!!!!!!!!!!!