Answer:
(a): The magnitude of the electric force on the small sphere = ![\dfrac{q\sigma}{2\epsilon_o}.](https://tex.z-dn.net/?f=%5Cdfrac%7Bq%5Csigma%7D%7B2%5Cepsilon_o%7D.)
(b): Shown below.
Explanation:
<u>Given:</u>
- m = mass of the small sphere.
- q = charge on the small sphere.
- L = length of the silk fiber.
= surface charge density of the large vertical insulating sheet.
<h2>
(a):</h2>
When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:
![\rm E = \dfrac{\sigma}{2\epsilon_o}.](https://tex.z-dn.net/?f=%5Crm%20E%20%3D%20%5Cdfrac%7B%5Csigma%7D%7B2%5Cepsilon_o%7D.)
<em>where,</em>
is the electrical permittivity of the free space.
The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:
![\rm F_e=qE.](https://tex.z-dn.net/?f=%5Crm%20F_e%3DqE.)
Thus, the magnitude of the electric force on the small sphere is given by
![\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.](https://tex.z-dn.net/?f=%5Crm%20F_e%20%3D%20q%5Ctimes%20%5Cdfrac%7B%5Csigma%20%7D%7B2%5Cepsilon_o%7D%3D%5Cdfrac%7Bq%5Csigma%7D%7B2%5Cepsilon_o%7D.)
The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.
<h2>
(b):</h2>
When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.
According to the fig.,
![\rm \tan \theta = \dfrac{F_e}{W}.](https://tex.z-dn.net/?f=%5Crm%20%5Ctan%20%5Ctheta%20%3D%20%5Cdfrac%7BF_e%7D%7BW%7D.)
<em>where,</em>
= electric force on the sphere, acting along left.
= weight of the sphere, acting vertically downwards.
<em />
![\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .](https://tex.z-dn.net/?f=%5Crm%20F_e%20%3D%20%5Cdfrac%7Bq%5Csigma%7D%7B2%5Cepsilon_o%7D%5C%5C%5C%5CW%3Dmg%5C%5C%5C%5CTherefore%2C%5C%5C%5C%5C%5Ctan%5Ctheta%20%3D%20%5Cdfrac%7B%5Cdfrac%7Bq%5Csigma%7D%7B2%5Cepsilon_o%7D%7D%7Bmg%7D%3D%5Cdfrac%7Bq%5Csigma%7D%7B2mg%5Cepsilon_o%7D.%5C%5C%5CRightarrow%20%5Ctheta%3D%5Ctan%5E%7B-1%7D%5Cleft%20%28%20%5Cdfrac%7Bq%5Csigma%7D%7B2mg%5Cepsilon_o%7D%5Cright%20%29%20.)
g is the acceleration due to gravity.
Given:
m = 4 kg, the mass of the object
h = 5 m, distance fallen
Neglect air resistance.
The PE (potential energy) is
PE = mgh = (4 kg)*(9.8 m/s²)*(5 m) = 196 J
The PE is converted into KE (kinetic energy) after the fall.
Therefore the PE decreased by 196 J ≈ 200 J
Answer: d. It has decreased by 200 J
I would say B because it is near the ocean which can cause a tsunami but also because of the wind coming from the ocean (it might cause hurricanes and lots of storms) I’m not sure though but that’s what I think makes sense. Good Luck!
Dont click on the link its a scam btw
Answer:
8.2 m/s²
Explanation:
m = mass of the block
μ = Coefficient of kinetic friction = 0.17
= Normal force on the block by the ramp
= kinetic frictional force
Force equation perpendicular to ramp surface is given as
![F_{n} = mg Cos65](https://tex.z-dn.net/?f=F_%7Bn%7D%20%3D%20mg%20Cos65)
Kinetic frictional force is given as
![f_{k} = \mu F_{n}](https://tex.z-dn.net/?f=f_%7Bk%7D%20%3D%20%5Cmu%20F_%7Bn%7D)
![f_{k} = \mu mg Cos65](https://tex.z-dn.net/?f=f_%7Bk%7D%20%3D%20%5Cmu%20mg%20Cos65)
Force equation parallel to ramp surface is given as
![mg Sin65 - f_{k} = ma](https://tex.z-dn.net/?f=mg%20Sin65%20-%20f_%7Bk%7D%20%3D%20ma)
![mg Sin65 - \mu mg Cos65 = ma](https://tex.z-dn.net/?f=mg%20Sin65%20-%20%5Cmu%20mg%20Cos65%20%3D%20ma)
![g Sin65 - \mu g Cos65 = a](https://tex.z-dn.net/?f=g%20Sin65%20-%20%5Cmu%20g%20Cos65%20%3D%20a)
![(9.8) Sin65 - (0.17) (9.8) Cos65 = a](https://tex.z-dn.net/?f=%289.8%29%20Sin65%20-%20%280.17%29%20%289.8%29%20Cos65%20%3D%20a)
m/s²