Answer:

Explanation:
Let consider the following system, which is described in the image attached below and two reference axis, one parallel and the other perpendicular to the direction of motion. The corresponding equations of equilibrium are described herein:


The acceleration of the bowling ball at the lowest point occurs at 



Since no external torque is acting on the system you can use the conservation of angular momentum. I derived the final angular speed below and shown my work on how I did it. It’s now just a matter of plugging in the numbers and using correct placement of negative sign for direction of angular velocity. L in the picture stands for angular momentum. Hope it helps
Answer:
Answer:
Explanation:
Given that
K=8.98755×10^9Nm²/C²
Q=0.00011C
Radius of the sphere = 5.2m
g=9.8m/s²
1. The electric field inside a conductor is zero
εΦ=qenc
εEA=qenc
net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero
This surface encloses no charge, and thus qenc=0. Gauss’ law.
Since it is inside the conductor
E=0N/C
2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as
F=kq1/r²
F=kQ/r²
F=8.98755E9×0.00011/5.2²
F=36561.78N/C
The electric field at the surface of the conductor is 36561N/C
Since the charge is positive the it is outward field
3. Given that a test charge is at 12.6m away,
Then Electric field is given as,
E=kQ/r²
E=8.98755E9 ×0.00011/12.6²
E=6227.34N/C
Explanation:
Given that,
Distance 1, r = 100 m
Intensity, 
If distance 2, r' = 25 m
We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :
.........(1)
Let I' is the intensity at r'. So,
............(2)
From equation (1) and (2) :



Intensity level is given by :
, 

dB = 32.96 dB
Hence, this is the required solution.
True. Since Inertia states this law, we are bound to believe greater mass = greater resistance in change of motion