Answer:
The electric flux is 
Explanation:
Given:
- Radius of the disc R=0.50 m
- Angle made by disk with the horizontal
- Magnitude of the electric Field
The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows
where
is the total Electric Flux- E is the Electric Field
- dA is the Area through which the electric flux is to be calculated.
Now according to question we have
Hence the electric flux is calculated.
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Force, pressure, and charge are all what are called <em>derived units</em>. They come from algebraic combinations of <em>base units</em>, measures of things like length, time, temperature, mass, and current. <em>Speed, </em>for instance, is a derived unit, since it's a combination of length and time in the form [speed] = [length] / [time] (miles per hour, meters per second, etc.)
Force is defined with Newton's equation F = ma, where m is an object's mass and a is its acceleration. It's unit is kg·m/s², which scientists have called a <em>Newton</em>. (Example: They used <em>9 Newtons</em> of force)
Pressure is force applied over an area, defined by the equation P = F/A. We can derive its from Newtons to get a unit of N/m², a unit scientists call the <em>Pascal</em>. (Example: Applying <em>100 Pascals </em>of pressure)
Finally, charge is given by the equation Q = It, where I is the current flowing through an object and t is how long that current flows through. It has a unit of A·s (ampere-seconds), but scientist call this unit a Coulomb. (Example: 20 <em>Coulombs</em> of charge)
Explanation:
Formula for steady flow energy equation for the flow of fluid is as follows.
![m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w](https://tex.z-dn.net/?f=m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%5D%20%2B%20z_%7B1%7Dg%5D%20%2B%20q%20%3D%20m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%20%2B%20z_%7B1%7Dg%5D%20%2B%20w)
Now, we will substitute 0 for both 
and 
, 0 for w, 334.9 kJ/kg for 
, 2726.5 kJ/kg for 
, 5 m/s for 
and 220 m/s for 
.
Putting the given values into the above formula as follows.
![1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0](https://tex.z-dn.net/?f=1%20%5Ctimes%20%5B334.9%20%5Ctimes%2010%5E%7B3%7D%20J%2Fkg%20%2B%20%5Cfrac%7B%285%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%20q%20%3D%201%20%5Ctimes%20%5B2726.5%20%5Ctimes%2010%5E%7B3%7D%20%2B%20%5Cfrac%7B%28220%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%200)
q = 6597.711 kJ
Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.
