Answer:
The friction force and the x component for the weight should be the reaction forces that are opposite and equal to the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.
Explanation:
<u>When the locomotive starts to pull the train up, appears two reaction forces opposed to the action force in the direction of the move. </u>
The first one is due to the friction between the wheels and the ground, it will be the friction force (Fr):
Fr = μ*Pₓ =μmg*sin(φ)
<em>where μ: friction dynamic coefficient, Pₓ: is the weight component in the x-axis, m: total mass = train's mass + locomotive's mass, g: gravity, and sin(φ): is the angle respect to the x-axis.</em>
And the second one is the x component for the weight (Wₓ):
Wₓ = mg*cos(φ)
<em>where cos(φ): is the angle respect to the y-axis. </em>
<em> </em>
These two forces should be the same as the action force, which causes the locomotive to move up the hill if the velocity of the locomotive remains constant.
Answer:
The average emf induced in the coil is 175 mV
Explanation:
Given;
number of turns of the coil, N = 1060 turns
diameter of the coil, d = 20.0 cm = 0.2 m
magnitude of the magnetic field, B = 5.25 x 10⁻⁵ T
duration of change in field, t = 10 ms = 10 x 10⁻³ s
The average emf induced in the coil is given by;
where;
A is the area of the coil
A = πr²
r is the radius of the coil = 0.2 /2 = 0.1 m
A = π(0.1)² = 0.03142 m²
Therefore, the average emf induced in the coil is 175 mV
Answer:
speed =distance/time taken
5 m/s
Answer:
<em>126.01 rad/s^2</em>
<em></em>
Explanation:
since it starts from rest, initial angular speed ω' = 0 rad/s
angular speed N = 477 rev/min
angular speed in rad/s ω = 
= 
= 49.95 rad/s
angular displacement ∅ = 1.5758 rev
angular displacement in rad/s = 
= 2 x 3.142 x 1.5758 = 9.9 rad
angular acceleration 
= ?
using the equation of angular motion
ω^2 = ω'^2 + 2∅
imputing values, we have
2495 = 19.8
= 2495/19.8 = <em>126.01 rad/s^2</em>
