Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
Answer:
The energy stored is: 62.5 Joules
Explanation:
Given
--- spring constant
--- stretch
Required
The amount of energy
This is calculated as:
In the above problem, we need to find mass of the second child, so that the Center of Mass remains at the origin( pivot).
CM= m1r1+m2r2/m1+m2
0= 20*-2+16*r2/20+16
r2= 40/16
r2= +2.5 m
Answer:
cross at many points
Explanation:
According to the diagram it does not shows the real electric field as the electric field does not intersect with each other but according to the given diagram many lines are intersected with each other
So the correct option is C as it would be crossed at many points
Hence, the same would be relevant
