1) ![1.11\cdot 10^{-7} J](https://tex.z-dn.net/?f=1.11%5Ccdot%2010%5E%7B-7%7D%20J)
The capacitance of a parallel-plate capacitor is given by:
![C=\frac{\epsilon_0 A}{d}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B%5Cepsilon_0%20A%7D%7Bd%7D)
where
is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is
![r=\frac{2.0 cm}{2}=1.0 cm=0.01 m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B2.0%20cm%7D%7B2%7D%3D1.0%20cm%3D0.01%20m)
so the area is
![A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%280.01%20m%29%5E2%3D3.14%5Ccdot%2010%5E%7B-4%7D%20m%5E2)
While the separation between the plates is
![d=0.50 mm=5\cdot 10^{-4} m](https://tex.z-dn.net/?f=d%3D0.50%20mm%3D5%5Ccdot%2010%5E%7B-4%7D%20m)
So the capacitance is
![C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B%288.85%5Ccdot%2010%5E%7B-12%7D%20F%2Fm%29%283.14%5Ccdot%2010%5E%7B-4%7D%20m%5E2%29%7D%7B5%5Ccdot%2010%5E%7B-4%7D%20m%7D%3D5.56%5Ccdot%2010%5E%7B-12%7D%20F)
And now we can find the energy stored,which is given by:
![U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7DCV%5E2%3D%5Cfrac%7B1%7D%7B2%7D%285.56%5Ccdot%2010%5E%7B-12%7D%20F%2Fm%29%28200%20V%29%5E2%3D1.11%5Ccdot%2010%5E%7B-7%7D%20J)
2) 0.71 J/m^3
The magnitude of the electric field is given by
![E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BV%7D%7Bd%7D%3D%5Cfrac%7B200%20V%7D%7B5%5Ccdot%2010%5E%7B-4%7D%20m%7D%3D4%5Ccdot%2010%5E5%20V%2Fm)
and the energy density of the electric field is given by
![u=\frac{1}{2}\epsilon_0 E^2](https://tex.z-dn.net/?f=u%3D%5Cfrac%7B1%7D%7B2%7D%5Cepsilon_0%20E%5E2)
and using
, we find
![u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3](https://tex.z-dn.net/?f=u%3D%5Cfrac%7B1%7D%7B2%7D%288.85%5Ccdot%2010%5E%7B-12%7D%20F%2Fm%29%284%5Ccdot%2010%5E5%20V%2Fm%29%5E2%3D0.71%20J%2Fm%5E3)
Answer:
kinetic friction may be greater than 400 N or smaller than 400 N
Explanation:
As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as
![F_s = \mu_s N](https://tex.z-dn.net/?f=F_s%20%3D%20%5Cmu_s%20N)
now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as
![F_k = \mu_k N](https://tex.z-dn.net/?f=F_k%20%3D%20%5Cmu_k%20N)
now we know that
![\mu_k < \mu_s](https://tex.z-dn.net/?f=%5Cmu_k%20%3C%20%5Cmu_s)
so here value of limiting static friction force is always more than kinetic friction
also we know that
initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction
and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction
so kinetic friction may be greater than 400 N or smaller than 400 N
The stopwatch will be the most useful in determining the kinetic energy of a 50 g battery- powered car traveling a distance of 10 m.
<h3>What is kinetic energy?</h3>
Kinetic energy is the energy of a body possessed due to motion.
This means that for an object to possess kinetic energy, it must be in motion.
The kinetic energy is measured in Joules, which is a product of the mass of the substance and the time taken to travel a distance.
A stopwatch is an instrument used to measure time as one of the components of kinetic energy.
Therefore, the stopwatch will be the most useful in determining the kinetic energy of a 50 g battery- powered car traveling a distance of 10 m.
Learn more about kinetic energy at: brainly.com/question/12669551
Answer:
The differences that will be observed are;
1) The Sun will become faint and will no longer be yellow but rather appear white and will no longer be visible (become invisible) by unassisted vision as we can see the Sun today
2) The size of the Sun will shrink to a size comparable to the size of the Earth
3) The Sun will cool down and will no longer radiate as much heat
4) The nuclear reactions that generate energy on the Sun's will seize and the and the heat from the Sun will be from residual thermal energy
5) The core, which is the hottest part of the Sun will no longer be hydrogen but carbon and oxygen
Explanation: