Ep= mgh
Ep = 40 x 9.8 x 10
Ep = 3920J
Ep = 3900J (2sf)
Answer:
<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>

</h2>
Explanation:
Part a)
As we know that drag force is given as






so we have


So acceleration of the ball is



Part B)
As per kinematics we know that



Answer:
Current in outer circle will be 15.826 A
Explanation:
We have given number of turns in inner coil 
Radius of inner circle 
Current in the inner circle 
Number of turns in outer circle 
Radius of outer circle 
We have to find the current in outer circle so that net magnetic field will zero
For net magnetic field current must be in opposite direction as in inner circle
We know that magnetic field is given due to circular coil is given by

For net magnetic field zero

So 

Answer:
10250 N/C leftwards
Explanation:
QA = 4 micro Coulomb
QB = - 5 micro Coulomb
AP = 6 m
BP = 2 m
A is origin, B is at 4 m and P is at 6 m .
The electric field due to charge QA at P is EA rightwards

The electric field due to charge QB at P is EB leftwards

The resultant electric field at P due the charges is given by
E = EB - EA
E = 11250 - 1000 = 10250 N/C leftwards
Answer:
Element Is The Answer I think