Answer:
298rad/s and 116.96 ohms
Explanation:
Given an L-R-C series circuit where
L = 0.450 H,
C=2.50×10^−5F, and resistance R= 0
In this situation we have a simple LC circuit with angular frequency
Wo = 1√LC
= 1/√(0.450)(2.50×10^-5)
= 1/√0.00001125
= 1/0.003354
= 298rad/s
B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.
Wi/W° = (100-10)/100
Wi/W° = 90/100
Wi/W° = 0.90 ............... 1
Angular frequency of oscillation
The complete aspect of the solution is attached, please check.
Answer: 1.3 *10^6 Ω*m
Explanation: In order to explain this problem we have to use the following expression for the resistence:
R=L/(σ*A) where L and A are the length and teh area for the wire, respectively. σ is the conductivity of teh Nichrome.
Then, from mteh OHM law we have V=R*I so R=V/I=2/3.2=0.625 Ω
Finally we have:
σ=L/(R*A)=1.3/(0.625*1.6*10^-6)=1.3*10^6 Ω*m
MY personal interpretation of nothing is no atoms or particles of anything. but keep in mind im 11 <span />
Explanation:
As the given spheres are connected by a thin wire so, the potential on the spheres are the same.
......... (1)
Hence, total charge will be as follows.
= Q = -95.5 nC .......... (2)
Using the above two equations, the final equation will be as follows.
and, 
Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.
= 
= 82.714 nC
Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.
Answer:
That an item is neither moving nor staying still in a position that is building up energy.
Explanation:
