Answer:
The magnitude of the induced Emf is
Explanation:
The width of the truck is given as 79inch but we need to convert to meter for consistency, then
The width= 79inch × 0.0254=2.0066 metres.
Now we can calculate the induced Emf using expresion below;
Then the
Where B= magnetic field component
L= width
V= velocity
=(40*10^-6) × (42) × (2.0066)
=0.003371V
Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V
Answer:
-22.2 m/s²
Explanation:
The equation for position x for a constant acceleration a, time t and initial velocity v₀, initial position x₀:
(1)
For rocket A the initial and final position: x = x₀= 0. Using these values in equation 1 gives:
(2)
Solving for time t:
(3)
The times for both rockets must be equal, since they start and end at the same location. Using equation 3 for rocket A and B gives:
(4)
Solving equation 4 for acceleration of rocket B:
(5)
Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
