Answer:
Explanation:
F = ma
4.45g - 2.75g = (4.45 + 2.75)a
a = 9.81(4.45 - 2.75) / (4.45 + 2.75) = 2.31625 ≈ 2.32 m/s²
a)
T = 2.75(9.81 + 2.32) = 33.3 N
or
T = 4.45(9.81 - 2.32) = 33.3 N
b) 2.32 m/s² upward for 2.75 kg mass
2.32 m/s² downward for 4.45 kg mass
c) y = ½at² = ½(2.31625/3)1² = 1.158125 ≈ 1.16 m
The direction of work.........
Answer:


Explanation:
From the question we are told that
Initial velocity of 60 m/s
Wind speed 
Generally Resolving vector mathematically

Generally the equation Pythagoras theorem is given mathematically by



Therefore Resultant velocity (m/s)

b)Resultant direction
Generally the equation for solving Resultant direction

Therefore


Answer:
A the type of dog food because it was different and every other thing was the same.
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