Answer:
Given:
radius of the coil, R = 6 cm = 0.06 m
current in the coil, I = 2.65 A
Magnetic field at the center, B =
Solution:
To find the number of turns, N, we use the given formula:

Therefore,

N = 22.74 = 23 turns (approx)
Answer:
m = 56.5 kg
Explanation:
Since the addition of mass on one piston caused a change in pressure head at the other. Diameter of the piston calculated is used as 0.46 m
Δm*g / Area = p * g * Δh ..... Eq1


Answer:
The minimum wall thickness required for the spherical tank is 0.0189 m
Explanation:
Given data:
d = inside diameter = 8.1 m
P = internal pressure = 1.26 MPa
σ = 270 MPa
factor of safety = 2
Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?
The allow factor of safety:

The minimun wall thickness:

Ok bro you are going to have 1 more