Answer:
at y=6.29 cm the charge of the two distribution will be equal.
Explanation:
Given:
linear charge density on the x-axis, 
linear charge density of the other charge distribution, 
Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.
Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.
<u>we know, the electric field due to linear charge is given as:</u>
where:
linear charge density
r = radial distance from the center of wire
permittivity of free space
Therefore,
∴at y=6.29 cm the charge of the two distribution will be equal.
The sound is increased because sound waves are in fact mech. waves which means the that they can't travel through empty space and thus need a medium to travel through
Answer: m= 35.6 kg
Explanation:
For finding the mass of the stone we have the formula
v= 
Here, Tension= m*g = m*9.81
and linear mass density= 
Linear mass density= 
Linear mass density= 0.0127 kg/m
Velocity= 
Velocity= 2 * 
Velocity= 165.8 m/s
So putting all these values in equation we get
v=
165.8= 
Solving we get
m= 35.58 kg
or m= 35.6 kg
Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
