Efficiency = useful energy out / total energy in x 100
= 100/400 x 100
=0.25 x 100
= 25%
25%
Answer:
The induced current can be increased in the coil in the following ways: By increasing the strength of the magnet. By increasing the speed of the magnet through the coil.
Explanation:
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
<span>Standard deviation is a calculation. It I used in statistical analysis of a easier job. hoped this helps u </span>
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.