the force experienced by an electron in a field between parallel plates is proportional to which of the following?
1 answer:
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Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
Answer:
71.76 m
Explanation:
We will solve this question using the work energy theorem.
The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.
ΔK.E = W
In the attached free body diagram for the question, the forces acting on the puck are given.
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Final kinetic energy = 0 J (since the puck comes to a stop)
Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J
ΔK.E = 0 - 67.6 = - 67.6 J
W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)
Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h
Workdone by the frictional force = F × d
F = μ N
μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)
N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N
F = μ N = 0.3 × 1.697 = 0.509 N
where d = distance along the incline that the puck travels.
d = h/sin 30° = 2h (from trigonometric relations)
Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h
ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)
- 67.6 = - 1.96h + 1.02h
-0.942h = - 67.6
h = 71.76 m
