Answer:
well the answers are 180 and 240 obviously
Explanation:
In electricity, the most famous and basic equation is the Ohm's Law which relates the parameters voltage, current and resistance. One form of this law as written in equation is V = IR, where V is the voltage in volts, I is the current in amperes and R is the resistance in ohms. These parameters depends in the arrangements, whether it's series or parallel.
In a series connection, the voltage is greater across a high-resistance resistor. Therefore, the voltage is much greater for the 20-ohm resistor. However,if it is a parallel circuit, the voltage is just the same for both resistors.
A bowler who always left the same 3 pins standing could be considered a C. Precise bowler as from bowling countless number of times he has observed the same amount of pins knocked down each time.
Answer:
The friction force acting on the object is 7.84 N
Explanation:
Given;
mass of object, m = 4 kg
coefficient of kinetic friction, μk = 0.2
The friction force acting on the object is calculated as;
F = μkN
F = μkmg
where;
F is the frictional force
m is the mass of the object
g is the acceleration due to gravity
F = 0.2 x 4 x 9.8
F = 7.84 N
Therefore, the friction force acting on the object is 7.84 N
Question:
A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
Explanation:
A point charge ,q = 
is located in the center of a spherical cavity of radius , 
m inside an insulating spherical charged solid.
The charge density in the solid , d = 
Distance from the center of the cavity,R =
Volume of shell of charge= V =![(\frac{4\pi}{3})[ R^3 - r^3 ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D)
Charge on the shell ,Q = 
![Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d](https://tex.z-dn.net/?f=Q%20%3D%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D%20%5Ctimes%20d)
![Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%2A%5Ctimes%2010%5E%7B-4%20%7D%5B8.57375%20-%202.81011%20%5D%5Ctimes%207.35%5Ctimes%2010%5E%7B-4%7D)
![Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%5Ctimes%2010%5E%7B-4%7D%20%5B5.76364%20%5D%20%5Ctimes%207.35%20%5Ctimes%2010%5E%7B-4%7D)
Electric field at 
m due to shell
E1 = 
Electric field at due to 'q' at center 
E2 =
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
![=[ 2.134 - 1.769 ]\times 10^6](https://tex.z-dn.net/?f=%3D%5B%20%202.134%20%20-%201.769%20%5D%5Ctimes%2010%5E6)
