Answer:
4. The reading decreases because part of your weight is applied to the table and the table exerts a matching force on you that acts in the direction opposite to your weight.
Explanation:
When you push on anything, Newtons Third Law comes into effect: "when two objects interact, they apply forces to each other of equal magnitude and opposite direction." So pretty much, the table you are putting some of your weight on is taking some of the downward force and in return the scale isn't going to push up with as much force.
I'm not the best at explaining, but im in physics right now and thats somewhat how my teacher explained it.
Explanation:
pressure=force/area
force=mass x acceleration
area=length x width
force=M x LT^-2
area=L^2
![\frac{mlt { - }^{2} }{l {}^{2} }](https://tex.z-dn.net/?f=%20%20%5Cfrac%7Bmlt%20%7B%20-%20%7D%5E%7B2%7D%20%7D%7Bl%20%7B%7D%5E%7B2%7D%20%7D%20)
=ML^-1T^-2
To solve this problem we will apply the concepts related to energy conservation. We will start by defining the expressions of the electric potential energy for a given charge (and for the electron). With this we can apply the conservation of kinematic energy. Our values are given as
![V = 80000V](https://tex.z-dn.net/?f=V%20%3D%2080000V)
The potential energy:
![U = qV](https://tex.z-dn.net/?f=U%20%3D%20qV)
Here,
q = Charge
V = Voltage
Or specifically for an electron we can define it as,
![U_e = eV](https://tex.z-dn.net/?f=U_e%20%3D%20eV)
Here,
e = Charge of electron
V = Voltage
Applying the energy conservation equations we have that the kinetic energy must be equivalent to the electric potential energy,
![KE= U_e](https://tex.z-dn.net/?f=KE%3D%20U_e)
![\frac{1}{2}mv^2 = qV =e V](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20qV%20%3De%20V)
Here
v = Velocity
m = Mass
Rearranging,
![v^2 = \frac{2eV}{m}](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7B2eV%7D%7Bm%7D)
Replacing,
![v^2 = \frac{2(1.6*10^{-19})(80000)}{9.1*10^{-31}}](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7B2%281.6%2A10%5E%7B-19%7D%29%2880000%29%7D%7B9.1%2A10%5E%7B-31%7D%7D)
![v= 2.81*10^{16}](https://tex.z-dn.net/?f=v%3D%202.81%2A10%5E%7B16%7D)
For each electron the velocity is,
![v = 1.68*10^8m/s](https://tex.z-dn.net/?f=v%20%3D%201.68%2A10%5E8m%2Fs)
Therefore the velocity of the electron is ![1.68*10^8m/s](https://tex.z-dn.net/?f=1.68%2A10%5E8m%2Fs)
15 = a+(3-1)d
23 = a+(7-1)d
simultaneous eqs. solve for a and d
(a) If the balloon is suspended in air and it drifts neither up nor down, it means the forces acting on it are equally balanced. There are only two forces acting on the balloon:
- its weight, of magnitude 1 N, directed downward
- the buoyant force, directed upward
The two forces are balanced, so the buoyant force in this case is equal to the weight of the balloon: 1 N.
(b) when the buoyant force decreases, the weight becomes greater than the buoyant force. This means there is now a net force on the balloon, acting downward, so the balloon starts to drift toward the ground.
(c) when the buoyant force increases, it becomes greater than the weight. This means there is now a net force on the balloon, acting upward, so the balloon starts to go up.