Answer and Explanation:
The main source of earth magnetic field is the metals and iron present in earth crust these metals are mainly found in liquid state in earth crust and we know that whenever there is spinning of these liquid metals then there will magnetic field generate, So the main source of earth magnetic field is the metals present in earth crust.
F=W=mg
12N(given)= m*9.8
m= 1.22 kg
A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is
43.8 N = <em>k</em> (0.155 m) ==> <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.
Answer:
The net Electric field at the mid point is 289.19 N/C
Given:
Q = + 71 nC = 
Q' = + 42 nC = 
Separation distance, d = 1.9 m
Solution:
To find the magnitude of electric field at the mid point,
Electric field at the mid-point due to charge Q is given by:



Now,
Electric field at the mid-point due to charge Q' is given by:



Now,
The net Electric field is given by:

