Answer:
The permittivity of rubber is 
Explanation:
From the question we are told that
The magnitude of the point charge is 
The diameter of the rubber shell is 
The Electric field inside the rubber shell is 
The radius of the rubber is mathematically evaluated as
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
Where 
is the permittivity of rubber
=> 
=> 
substituting values
Answer:
19.1 deg
Explanation:
v = speed of the proton = 8 x 10⁶ m/s
B = magnitude of the magnetic field = 1.72 T
q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C
F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N
θ = Angle between proton's velocity and magnetic field
magnitude of magnetic force on the proton is given as
F = q v B Sinθ
7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ
Sinθ = 0.327
θ = 19.1 deg
A. the medium through which the light travels changes.
Explanation:
Light waves will continue to travel in a straight line in all directions from their source unless the medium through which the light travels changes.
A change in medium causes light to exhibit different properties. Also, when light hits an obstacle, they can be diffracted.
- The way light travels on crossing a boundary differs.
- At the boundary between two medium, light can either be reflected back or refracted when they cross the medium
- This will cause the light rays to bend towards or away from the normal depending on the properties of the medium.
Learn more:
Refraction brainly.com/question/12370040
#learnwithBrainly
Answer:
I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL
C IS CORRECT
Answer:
The distance will be x = 41.7 [m]
Explanation:
We must first find the components in the x & y axes of the initial velocity.
![(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]](https://tex.z-dn.net/?f=%28v_%7Bo%7D%29_%7Bx%7D%20%3D%2015%2Acos%2820%29%3D%2014.09%5Bm%2Fs%5D%5C%5C%28v_%7Bo%7D%29_%7By%7D%20%3D%2015%2Asin%2820%29%3D%205.13%5Bm%2Fs%5D)
The acceleration is the gravity acceleration therefore.
g = 9.81 [m/s^2]
Now we can calculate how long it takes to fall.
![y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]](https://tex.z-dn.net/?f=y%3D%28v_%7Bo%7D%29_%7By%7D%2At-0.5%2Ag%2At%5E2%5C%5C-28%20%3D%205.13%2At-0.5%2A9.81%2At%5E2%5C%5C-28%3D-4.905%2At%5E2%2B5.13%2At%5C%5C4.905%2At%5E2-5.13%2At%3D28%5C%5Ct%20%3D%202.96%5Bs%5D)
With this time we can find the horizontal distance that runs the projectile.
![x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bo%7D%29_%7Bx%7D%2At%5C%5Cx%3D14.09%2A2.96%5C%5Cx%3D41.7%5Bm%5D)
