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The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
i = 0.477 10⁴ B
the current flows in the counterclockwise
Explanation:
For this exercise let's use the Ampere law
∫ B . ds = μ₀ I
Where the path is closed
Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.
From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.
We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field
Segment on the y axis
L₀ = (y2-y1)
L₀ = 3-0 = 3 cm
Segment on the point x = 2 cm
L₁ = 3-0
L₁ = 3cm
B L = μ₀ I
B 2L = μ₀ I
i = 2 L B /μ₀
i= 2 0.03 / 4π 10⁻⁷ B
i = 4.77 10⁴ B
The current is perpendicular to the magnetic field whereby the current flows in the counterclockwise
Answer: The minimum acceleration for the air plane is 2.269m/s2.
Explanation: To solve such problem the equation of motion are applicable.
The initial velocity is 0 since the airplane was initially standing. We are going to use this equation
V^2=U^2+2as
33^2=0+2a (240)
a= 2.269m/s2
