J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.
The rise in fluid level is 0.11 m, and for mercury, it is 0.728 cm or 7.28 mm. \nTo solve this, here’s the information we have: \n- Density of oil: [density value] \n- Change in pressure in the tank: [pressure change] \n- Density of mercury: [density value] \nTo find the fluid level rise in the manometer: \n1 mmHg equals 133.332 Pa. \nBased on the variables: g is the acceleration due to gravity and h represents the height of the fluid level. \nh = 0.11 m. \nUsing mercury, we find: \nh = 0.00728 m, which is 7.28 mm.
Answer:
A. 30.7 cm
B. 
C. The electric field points outward from the charge
Explanation:
A.
B.
Using Gauss's law
C. The electric field emanates away from the point of charge if the charge is positive.
