Answer:
N= 3
Explanation:
For this exercise we must use Faraday's law
E = - dФ / dt
Ф = B . A = B Acos θ
tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives
E = - A cos θ (B - B₀) / t
The angle enters the magnetic field and the normal to the area is zero
cos 0 = 1
A = π r²
In the length of the wire there are N turns each with a length L₀ = 2π r
L = N (2π r)
r = L / 2π N
we substitute
A = L² / (4π N²)
The magnetic field produced by a solenoid is
B = μ₀ N/L I
for which
B₀ = μ₀ N/L I
The final field is zero, because the current is zero
B = 0
We substitute
E = - (L² / 4π N²) (0 - μ₀ N/L I) / t
E = μ₀ L I / (4π N t)
N = μ₀ L I / (4π t E)
The electromotive force is E = 0.80 mV = 0.8 10⁻³ V
let's calculate
N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]
N = 320 10⁻⁷ / 9.6 10⁻⁶
N = 33.3 10⁻¹
N= 3
The northward components of the resultant displacement is 40.96 m and the westward components of the resultant displacement of the bird from its nest is 28.68 m.
<h3>
Displacement of the bird</h3>
The displacement of the bird is the change in the position of the bird.
<h3>Vertical component of the bird's displacement </h3>
Vy₁ = -25 m x sin(55)
Vy₁ = -20.48 m
Vy₂ = 75 m x sin(55)
Vy₂ = 61.44 m
Total vertical displacement = 61.44 m - 20.48 m = 40.96 m
<h3>Horizontal component of the bird's displacement </h3>
Vx₁ = -25 m x cos(55)
Vx₁ = -14.34 m
Vx₂ = 75 m x cos(55)
Vx₂ = 43.02 m
Total horizontal displacement = 43.02 m - 14.34 m = 28.68 m
Learn more about displacement here: brainly.com/question/2109763
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part 1
mass = ρ x V
mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg
PE (potential energy)= mgh
PE = 6608.2 kg x 9.81 x 403
PE = 2.61 x 10⁷ J
part 2
megaton of TNT (Mt) =4.2 x 10¹⁵ J
convert PE to Mt:
2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt
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V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0
V₂ (Vg)₂ = -WAy₂ = -20(9.81)(0.5) = -98.1J
The kinetic energy because the pool rotates about a fixed axis
W = VA/rA = VA/0.2 5VA
Mass momen of inertila about fixed axis which passes through point 0
I₀ = mko² = 50(0.280)² = 3.92 kg. m²
∴ The kinetic energy of the system is
T = 1/2 I₀w² + 1/2mAVA²
= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA²
Now that the system is at rest then T₁ = 0
Energy conservation is
T₁ +V₁ = T₂ + V₂
0+ 0 = 59VA² + (-98.1)
VA = 1.289 m/s
= 1.29 m/s
