The answer to the question can charging by friction occur only in solids is no. not at all. the static charge generated by following fluids can be hazardous. just as the static discharge when you touch a doorknob produces a spark, lightening is the spark from the static charge built up in the atmosphere.
Answer:
Weight and reaction.
Explanation:
With respect to Newton's third law of motion, action and reaction have equal magnitude but acts in opposite direction.
Thus, the brick of mass 10 kg has its weight (due to gravitational pull of the earth), which acts downwards through its center of mass. This makes the scale to produce an opposing force of equal magnitude, which acts upwards. These two forces acting in opposite directions balancing each other.
The two forces acting on the brick are: its weight and the reaction from the scale.
Answer:
8.0 s
Explanation:
The penny is in free fall, so its motion is a uniform accelerated motion with constant acceleration; therefore we can use the following suvat equation:
![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where
s is the displacement
u = 0 is the initial velocity of the penny
t is the time
is the acceleration of gravity
The displacement is
s = -311 m
So we can re-arrange the equation to solve for t, the time of flight:
![t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(-311)}{-9.8}}=8.0 s](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2s%7D%7Bg%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%28-311%29%7D%7B-9.8%7D%7D%3D8.0%20s)
Answer:
The current through the tube is 73.39A.
Explanation:
The relationship between the resistivity
, the electric field
, and the current density
is given by
![\rho = \dfrac{E}{J}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cdfrac%7BE%7D%7BJ%7D)
This equation can be solved for
to get:
![J = \dfrac{E}{\rho}](https://tex.z-dn.net/?f=J%20%3D%20%5Cdfrac%7BE%7D%7B%5Crho%7D)
Since the current is ![I = J\cdot A](https://tex.z-dn.net/?f=I%20%3D%20J%5Ccdot%20A)
![I= J\cdot A = \dfrac{E}{\rho} \cdot A](https://tex.z-dn.net/?f=I%3D%20J%5Ccdot%20A%20%20%3D%20%5Cdfrac%7BE%7D%7B%5Crho%7D%20%5Ccdot%20A)
Now, for the tube of mercury
,
, and the area is
; therefore,
![I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2](https://tex.z-dn.net/?f=I%3D%20%5Cdfrac%7B23N%2FC%7D%7B9.84%2A10%5E%7B-7%7D%5COmega%5Ccdot%20m%20%7D%20%2A3.14%2A10%5E%7B-6%7Dm%5E2)
![\boxed{I = 73.39A.}](https://tex.z-dn.net/?f=%5Cboxed%7BI%20%3D%2073.39A.%7D)
Hence, the current through the mercury tube is 73.39A.