Sound waves or bounces off the wall and light waves are waves of light
-1- was created in the 1600 by william gilbert
-2-When the charge is positive, electrons in the metal of the electroscope are attracted to the charge and move upward out of the leaves. This results in the leaves to have a temporary positive charge and because like charges repel, the leaves separate. When the charge is removed, the electrons return to their original positions and the leaves relax
3-
An electroscope is made up of a metal detector knob on top which is connected to a pair of metal leaves hanging from the bottom of the connecting rod. When no charge is present the metals leaves hang loosely downward. But, when an object with a charge is brought near an electroscope, one of the two things can happen.
Answer:
W = 290.7 dynes*cm
Explanation:
d = 1/5 cm = 0.2 cm
The force is in function of the depth x:
F(x) = 1000 * (1 + 2*x)^2
We can expand that as:
F(x) = 1000 * (1 + 4*x + 4x^2)
F(x) = 1000 + 4000*x + 4000*x^2
Work is defined as
W = F * d
Since we have non constant force we integrate

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2
W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3
W = 200 + 80 + 10.7 = 290.7 dynes*cm
Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2

Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;

T = 7.83 X10⁻⁷ s
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,

- Charge on the second charged particle,

- Position of the first charge =

- Position of the second charge =

The electric field at a point due to a charge
at a point
distance away is given by

where,
= Coulomb's constant, having value 
= position vector of the point where the electric field is to be found with respect to the position of the charge
.
= unit vector along
.
The electric field at the origin due to first charge is given by

is the position vector of the origin with respect to the position of the first charge.
Assuming,
are the units vectors along x and y axes respectively.

Using these values,

The electric field at the origin due to the second charge is given by

is the position vector of the origin with respect to the position of the second charge.

Using these values,

The net electric field at the origin due to both the charges is given by

Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.