Answer:
Energy stored in the capacitor is
Explanation:
It is given that,
Charge, ![q=1.5\ \mu C=1.5\times 10^{-6}\ C](https://tex.z-dn.net/?f=q%3D1.5%5C%20%5Cmu%20C%3D1.5%5Ctimes%2010%5E%7B-6%7D%5C%20C)
Potential difference, V = 36 V
We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :
![U=\dfrac{1}{2}q\times V](https://tex.z-dn.net/?f=U%3D%5Cdfrac%7B1%7D%7B2%7Dq%5Ctimes%20V)
U = 0.000027 J
![U=2.7\times 10^{-5}\ J](https://tex.z-dn.net/?f=U%3D2.7%5Ctimes%2010%5E%7B-5%7D%5C%20J)
So, the potential energy is stored in the capacitor is
. Hence, this is the required solution.
Before going to answer this question first we have to understand reflection and laws of reflection.
Reflection is the optical phenomenon in which light will bounce back to the same medium from which it had originated .
Whenever a light ray will incident on a mirror or any reflecting surface, it will be reflected. The ray which falls on the reflecting surface is called incident ray and the ray which is reflected is called reflected ray.
Let us consider a normal to the point of incidence.The angle made by incident ray with the normal is called angle of incidence.Let it be denoted as[ i ]
The angle made by the reflected ray with the normal is called angle of incidence.Let it be denoted as [r]
There are two types of reflection.One is called regular and other one is called as irregular.The laws of reflection is valid for both the types of reflection.
There are two laws of reflection.
FIRST LAW -It states that the incident ray,reflected ray and the normal to the point of incidence,all lie in one plane.
SECOND LAW- It states that that the angle of incidence is equal to the angle of reflection irrespective of the type of reflection.i.e i =r
Hence the correct answer will be angle of reflection.
Answer:
An apple, potato, and onion all taste the same if you eat them with your nose plugged
Explanation:
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as
![\Delta L=\dfrac{PL}{AE}](https://tex.z-dn.net/?f=%5CDelta%20L%3D%5Cdfrac%7BPL%7D%7BAE%7D)
![A=\dfrac{PL}{\Delta LE}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7BPL%7D%7B%5CDelta%20LE%7D)
![A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B6640%5Ctimes%20370%7D%7B0.53%5Ctimes%20110%5Ctimes%2010%5E3%7D)
A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm