Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = ![2.0\micro F = 2.0\times 10^{- 6} F](https://tex.z-dn.net/?f=2.0%5Cmicro%20F%20%3D%202.0%5Ctimes%2010%5E%7B-%206%7D%20F)
C' =
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:
![C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F](https://tex.z-dn.net/?f=C_%7Beq%7D%20%3D%20C%20%2B%20C%27%20%3D%202.0%20%2B%204.0%20%3D%206.0%5Cmicro%20F%20%3D%206.0%5Ctimes%2010%5E%7B-%206%7D%20F)
The energy of the capacitor, E is given by;
![E = \frac{1}{2}C_{eq}V^{2}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7DC_%7Beq%7DV%5E%7B2%7D)
![E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%206.0%5Ctimes%2010%5E%7B-%206%7D%5Ctimes%20300%5E%7B2%7D%20%3D%200.27%20J)
Complete Question
The complete question is shown on the first uploaded image
Answer:
The theoretical angular magnification lies within the angular magnification range
Explanation:
From the question we are told that
The focal length of B is ![f_{objective } = 43.0 \ cm](https://tex.z-dn.net/?f=f_%7Bobjective%20%7D%20%3D%20%2043.0%20%5C%20cm)
The focal length of A is ![f_{eye} = 10.4 \ cm](https://tex.z-dn.net/?f=f_%7Beye%7D%20%3D%20%2010.4%20%5C%20%20cm)
The theoretical angular magnification is mathematically represented as
![m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7Bf_%7Bobjective%20%7D%7D%7Bf_%7Beye%7D%7D%20%20%3D%20%20%5Cfrac%7B43.0%7D%7B10.4%7D)
![m = \frac{f_{objective }}{f_{eye}} = 4.175](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7Bf_%7Bobjective%20%7D%7D%7Bf_%7Beye%7D%7D%20%20%3D%20%204.175)
Form the question the measured angular magnification ranges from 4 -5
So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range
<span>Visible satellite images are like photos which are dependent on visible
light from the sun so they work best during the day. The sensor works by
detecting radiation within the range that wavelength is visible. Because of
this, the rays is usually seen as reaching earth from the East. </span>
Answer: hope it helps you...❤❤❤❤
Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.
No, not necessarily.
For instance, Newton’s 2nd law is F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.
But take a look at this (incorrect) equation for the force of gravity:
F=−G(m+M)Mm√|r|3r
It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.
It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.
A simpler counter example is 1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.
Thermal energy is the answer