i believ that the answer would be
the acceleration of B is 0.2
<span>Since there is no friction, conservation of energy gives change in energy is zero
Change in energy = 0
Change in KE + Change in PE = 0
1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0
1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0
(vf^2 - vi^2) = 2 x g x (hi - hf)
Since it starts from rest vi = 0
Vf = squareroot of (2 x g x (hi - hf))
For h1, no hf
Vf = squareroot of (2 x g x (hi - hf))
Vf = squareroot of (2 x 9.81 x 30)
Vf = squareroot of 588.6
Vf = 24.26
For h2
Vf = squareroot of (2 x 9.81 x (30 – 12))
Vf = squareroot of (9.81 x 36)
Vf = squareroot of 353.16
Vf = 18.79
For h3
Vf = squareroot of (2 x 9.81 x (30 – 20))
Vf = squareroot of (20 x 9.81)
Vf = 18.79</span>
The movements of the tectonic plates
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
