I think this must be if i remember The huge amounts of smoke and ash often seen billowing from active volcanoes generally travel vertically, carried upward by the powerful thermal updrafts volcanoes generate. There are two major types of pyroclastic flows. The first actually comes from the collapse of one of the typical columns of smoke and ash from a volcano. This is the fastest and most energetic type, caused when the weight of the material in the column is too much for the air pressure to support.
The other type of pyroclastic flow is caused by the collapse of a lava dome, the swelling of the earth caused by pressure from magma below. This pyroclastic flow actually has two major components, the visible ash cloud along with an avalanche of hot blocks from the disintegrating dome. The two types of flows leave different types of ash deposits, which geologists can identify long after an eruption.
Answer:
![W = 1.6 \times 10^{-17} J](https://tex.z-dn.net/?f=W%20%3D%201.6%20%5Ctimes%2010%5E%7B-17%7D%20J)
Explanation:
Work done to move a charge through a given potential difference is
![W = q(\Delta V)](https://tex.z-dn.net/?f=W%20%3D%20q%28%5CDelta%20V%29)
here
q = magnitude of charge
= potential difference
here it is given that charge of the particle is same as charge of proton
![q = 1.6 \times 10^{-19} C](https://tex.z-dn.net/?f=q%20%3D%201.6%20%5Ctimes%2010%5E%7B-19%7D%20C)
![\Delta V = 100 V](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20100%20V)
now we have
![W = (1.6 \times 10^{-19}) (100)](https://tex.z-dn.net/?f=W%20%3D%20%281.6%20%5Ctimes%2010%5E%7B-19%7D%29%20%28100%29)
![W = 1.6 \times 10^{-17} J](https://tex.z-dn.net/?f=W%20%3D%201.6%20%5Ctimes%2010%5E%7B-17%7D%20J)
Keona donates 1 ml of blood 473.176 times.
Each ml of blood has 1.06 gm of mass in it.
So the total mass of the blood is
(1.06 gm/ml) x (473.176 ml) = 501.566 grams
Rounded to the nearest hundredth, that's 501.57 grams.
<span>more lines = a lot of electrons returning back to ground state from same level</span>
Let the distance form origin of the point on x-axis be x.
then ![(-k*2*q/x)+(k*q/(2R-x))=0](https://tex.z-dn.net/?f=%28-k%2A2%2Aq%2Fx%29%2B%28k%2Aq%2F%282R-x%29%29%3D0)
solving for x,
we get
.
so, the curve is a sphere with center at ![(4R/3,0,0)](https://tex.z-dn.net/?f=%284R%2F3%2C0%2C0%29)
radius
.
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