Response:
(b) 10 Wb
Clarification:
Given;
angle of the magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
The equation for magnetic flux is;
Φ = BACosθ
where;
B denotes the magnetic field strength
A represents the area of the plane
θ is the inclination angle
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Next, calculate the magnetic flux through a 2.0 m² section of the same plane:
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
<pHence, the magnetic flux through a 2.0 m² area of the same plane is
10 Wb.Option "b"
<span>a. To determine the velocity at which the camera strikes the ground:
v^2 = (v0)^2 + 2ay = 0 + 2ay
v = sqrt{ 2ay }
v = sqrt{ (2)(3.7 m/s^2)(239 m) }
v = 42 m/s
The camera impacts the ground with a speed of 42 m/s.
b. To calculate the duration it takes for the camera to reach the bottom:
y = (1/2) a t^2
t^2 = 2y / a
t = sqrt{ 2y / a }
t = sqrt{ (2)(239 m) / 3.7 m/s^2 }
t = 11.4 seconds
The camera descends for 11.4 seconds before hitting the ground.</span>
Answer:
529.15 m/s
Explanation:
h = Highest point = 70000 m
g = Gravitational acceleration = 2 m/s²
m = Sulfur's mass
Since both potential and kinetic energies are conserved
The velocity at which the liquid sulfur exited the volcano is 529.15 m/s
