Answer:
(A) As it moves farther and farther from Q, its speed will keep increasing.
Explanation:
When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.
Mathematically:
![F=\frac{1}{4\pi.\epsilon_0} \times \frac{Q.q}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B1%7D%7B4%5Cpi.%5Cepsilon_0%7D%20%5Ctimes%20%5Cfrac%7BQ.q%7D%7Br%5E2%7D)
where:
r = distance between the charges
permittivity of free space
By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.
Answer:
The banking angle required is
.
Explanation:
Banking of a road is the act of constructing a road along a curved path at a certain angle to avoid skidding-off of vehicles plying it. Centripetal force is required to pull the object moving with a velocity 'v' towards the center of the curve for stability.
The velocity of a car navigating a banked road is given by:
v =
(rg ÷ tanθ)
where: r is the radius of the road, g is the gravitational force and θ is the banking angle.
⇒
= rg ÷ tanθ
tanθ = ![\frac{rg}{v^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7Brg%7D%7Bv%5E%7B2%7D%20%7D)
θ =
![\frac{rg}{v^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7Brg%7D%7Bv%5E%7B2%7D%20%7D)
=
(given that g = 10
)
=
![\frac{3049}{2079.36}](https://tex.z-dn.net/?f=%5Cfrac%7B3049%7D%7B2079.36%7D)
=
1.4663
θ = ![55.7^{0}](https://tex.z-dn.net/?f=55.7%5E%7B0%7D)
The banking angle required is
.
Hey there!
<span>The process by which hot and cold air are transferred in the atmosphere is convection.</span>