The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
Calculate the magnetic field strength at the ground. Treat the transmission line as infinitely long. The magnetic field strength is then given by:
B = μ₀I/(2πr)
B = magnetic field strength, μ₀ = magnetic constant, I = current, r = distance from line
Given values:
μ₀ = 4π×10⁻⁷H/m, I = 170A, r = 8.0m
Plug in and solve for B:
B = 4π×10⁻⁷(170)/(2π(8.0))
B = 4.25×10⁻⁶T
The earth's magnetic field strength is 0.50G or 5.0×10⁻⁵T. Calculate the ratio of the line's magnetic field strength to earth's magnetic field strength:
4.25×10⁻⁶/(5.0×10⁻⁵)
= 0.085
= 8.5%
The transmission line's magnetic field strength is 8.5% of that of earth's natural magnetic field. This is no cause for worry.
Answer:
a feat
Explanation:
cause the mf already 7ft on the dot there is no such thing as 6 ft 12
Positive Work.
Negative Work.
Case of zero work done.
Displacement at an angle to the force.
Energy.
Kinetic Energy.
work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement.
