-identifies an electric charge
-it can identify its polarity (positive or negative) if you compare it to a charge that you already know
-can identify the magnitude of a charge (how big of a charge it is)
Pressure
= Force/Area
Area = π(d^2)/4
= π(0.4^2)/4
=0.126 m2
Pressure
= 50/0.126
= 396.825 Pa

Hi pupil here's your answer ::
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Action and Reaction do not act on the same body !! If they acted on the same body, the resultant force will be zero and their could be never accelerated motion.
If both the forces acted on the same body, then if they are equal to opposite direction the object will remain stationary. If on of the forces is greater than other the object will move in the direction of greater force.
If both acted in the same direction there would be an accelrated motion.
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Hope this helps . . . . .
Answer:
A velocity time graph shows the change of velocity of an object with respect ot time. If the slope of the graph is increasing in the postive region, it means that the velocity is changing, if the slope is decreasing, it means the the velocity is decreasing, but the object is moving in the same direction (positve direction).
If this slope intersects the graph at x-axis, it means that the body has 0 velocity and has become still. After that, if the line enters in the negative region, it means that its velocity is started to increases again, but the body is movinging in the opposite direction (negative direction)
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